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Spring-mass systems practice problems

Spring-mass systems: find k and m from the setup, then the motion x(t). 60 problems with worked solutions; in 60 of them every equation is proved by a computer algebra system.

11 worked problems

Try each one before opening the solution. Every line is labeled with what checked it. How we verify

Problem 6.488

hard
A weight of \( \displaystyle 2 \) lb stretches a spring \( \displaystyle \frac{1}{2} \) ft. A damping force of \( \displaystyle \frac{1}{2} \) times the velocity acts on it. It starts \( \displaystyle \frac{1}{2} \) ft below equilibrium with velocity \( \displaystyle 1 \) ft/s (down is positive). Find the position x(t).
  1. \[ \frac{2}{1 \cdot \frac{1}{2}} = 4 \]
    Hooke's law: k = weight / stretch (lb/ft).✓ Proved
  2. \[ \frac{2}{32} = \frac{1}{16} \]
    m = weight / g, with g = 32 ft/s².✓ Proved
  3. The motion satisfies m x″ + c x′ + k x = 0: 1/16x″ + 1/2x′ + 4x = 0 (x measured downward from equilibrium).
    Reviewed
  4. \[ \left(\sqrt{3} \sin{\left(4 \sqrt{3} t \right)} + 2 \cos{\left(4 \sqrt{3} t \right)}\right) e^{- 4 t} + \frac{\frac{d}{d t} \frac{\left(\sqrt{3} \sin{\left(4 \sqrt{3} t \right)} + 2 \cos{\left(4 \sqrt{3} t \right)}\right) e^{- 4 t}}{4}}{2} + \frac{\frac{d^{2}}{d t^{2}} \frac{\left(\sqrt{3} \sin{\left(4 \sqrt{3} t \right)} + 2 \cos{\left(4 \sqrt{3} t \right)}\right) e^{- 4 t}}{4}}{16} = 0 \]
    This x(t) satisfies it ...✓ Proved
  5. \[ \left. \frac{\left(\sqrt{3} \sin{\left(4 \sqrt{3} t \right)} + 2 \cos{\left(4 \sqrt{3} t \right)}\right) e^{- 4 t}}{4} \right|_{\substack{ t=0 }} = \frac{1}{2} \]
    ... starts in the right place ...✓ Proved
  6. \[ \left. \frac{d}{d t} \frac{\left(\sqrt{3} \sin{\left(4 \sqrt{3} t \right)} + 2 \cos{\left(4 \sqrt{3} t \right)}\right) e^{- 4 t}}{4} \right|_{\substack{ t=0 }} = 1 \]
    ... with the right velocity.✓ Proved
Answer \( x(t) = \frac{\left(\sqrt{3} \sin{\left(4 \sqrt{3} t \right)} + 2 \cos{\left(4 \sqrt{3} t \right)}\right) e^{- 4 t}}{4} \)

✓ Nihil obstat Lines: 5 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 6.664

hard
A weight of \( \displaystyle 16 \) lb stretches a spring \( \displaystyle 2 \) ft. A damping force of \( \displaystyle 2 \) times the velocity acts on it. It starts \( \displaystyle \frac{1}{2} \) ft below equilibrium with velocity \( \displaystyle -2 \) ft/s (down is positive). Find the position x(t).
  1. \[ \frac{16}{2} = 8 \]
    Hooke's law: k = weight / stretch (lb/ft).✓ Proved
  2. \[ \frac{16}{32} = \frac{1}{2} \]
    m = weight / g, with g = 32 ft/s².✓ Proved
  3. The motion satisfies m x″ + c x′ + k x = 0: 1/2x″ + 2x′ + 8x = 0 (x measured downward from equilibrium).
    Reviewed
  4. \[ 2 \frac{d}{d t} \frac{\sqrt{3} e^{- 2 t} \cos{\left(2 \sqrt{3} t + \frac{\pi}{6} \right)}}{3} + \frac{\frac{d^{2}}{d t^{2}} \frac{\sqrt{3} e^{- 2 t} \cos{\left(2 \sqrt{3} t + \frac{\pi}{6} \right)}}{3}}{2} + \frac{8 \sqrt{3} e^{- 2 t} \cos{\left(2 \sqrt{3} t + \frac{\pi}{6} \right)}}{3} = 0 \]
    This x(t) satisfies it ...✓ Proved
  5. \[ \left. \frac{\sqrt{3} e^{- 2 t} \cos{\left(2 \sqrt{3} t + \frac{\pi}{6} \right)}}{3} \right|_{\substack{ t=0 }} = \frac{1}{2} \]
    ... starts in the right place ...✓ Proved
  6. \[ \left. \frac{d}{d t} \frac{\sqrt{3} e^{- 2 t} \cos{\left(2 \sqrt{3} t + \frac{\pi}{6} \right)}}{3} \right|_{\substack{ t=0 }} = -2 \]
    ... with the right velocity.✓ Proved
Answer \( x(t) = \frac{\sqrt{3} e^{- 2 t} \cos{\left(2 \sqrt{3} t + \frac{\pi}{6} \right)}}{3} \)

✓ Nihil obstat Lines: 5 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 6.486

hard
A \( \displaystyle 5 \) kg mass hangs from a spring with constant \( \displaystyle 80 \) N/m. A damping force of \( \displaystyle 40 \) times the velocity acts on it. It is released from rest \( \displaystyle \frac{1}{2} \) m below equilibrium. Find the position x(t).
  1. The motion satisfies m x″ + c x′ + k x = 0: 5x″ + 40x′ + 80x = 0 (x measured downward from equilibrium).
    Reviewed
  2. \[ 40 \left(4 t + 1\right) e^{- 4 t} + 40 \frac{d}{d t} \frac{\left(4 t + 1\right) e^{- 4 t}}{2} + 5 \frac{d^{2}}{d t^{2}} \frac{\left(4 t + 1\right) e^{- 4 t}}{2} = 0 \]
    This x(t) satisfies it ...✓ Proved
  3. \[ \left. \frac{\left(4 t + 1\right) e^{- 4 t}}{2} \right|_{\substack{ t=0 }} = \frac{1}{2} \]
    ... starts in the right place ...✓ Proved
  4. \[ \left. \frac{d}{d t} \frac{\left(4 t + 1\right) e^{- 4 t}}{2} \right|_{\substack{ t=0 }} = 0 \]
    ... with the right velocity.✓ Proved
Answer \( x(t) = \frac{\left(4 t + 1\right) e^{- 4 t}}{2} \)

✓ Nihil obstat Lines: 3 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 6.490

hard
A \( \displaystyle 4 \) kg mass hangs from a spring with constant \( \displaystyle 144 \) N/m. A damping force of \( \displaystyle 60 \) times the velocity acts on it. It starts \( \displaystyle \frac{1}{2} \) m below equilibrium with velocity \( \displaystyle -2 \) m/s (down is positive). Find the position x(t).
  1. The motion satisfies m x″ + c x′ + k x = 0: 4x″ + 60x′ + 144x = 0 (x measured downward from equilibrium).
    Reviewed
  2. \[ 8 \left(8 e^{9 t} + 1\right) e^{- 12 t} + 60 \frac{d}{d t} \frac{\left(8 e^{9 t} + 1\right) e^{- 12 t}}{18} + 4 \frac{d^{2}}{d t^{2}} \frac{\left(8 e^{9 t} + 1\right) e^{- 12 t}}{18} = 0 \]
    This x(t) satisfies it ...✓ Proved
  3. \[ \left. \frac{\left(8 e^{9 t} + 1\right) e^{- 12 t}}{18} \right|_{\substack{ t=0 }} = \frac{1}{2} \]
    ... starts in the right place ...✓ Proved
  4. \[ \left. \frac{d}{d t} \frac{\left(8 e^{9 t} + 1\right) e^{- 12 t}}{18} \right|_{\substack{ t=0 }} = -2 \]
    ... with the right velocity.✓ Proved
Answer \( x(t) = \frac{\left(8 e^{9 t} + 1\right) e^{- 12 t}}{18} \)

✓ Nihil obstat Lines: 3 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 6.491

hard
A \( \displaystyle 2 \) kg mass hangs from a spring with constant \( \displaystyle 50 \) N/m. A damping force of \( \displaystyle 10 \) times the velocity acts on it. It starts \( \displaystyle \frac{1}{4} \) m below equilibrium with velocity \( \displaystyle 1 \) m/s (down is positive). Find the position x(t).
  1. The motion satisfies m x″ + c x′ + k x = 0: 2x″ + 10x′ + 50x = 0 (x measured downward from equilibrium).
    Reviewed
  2. \[ \frac{5 \left(13 \sqrt{3} \sin{\left(\frac{5 \sqrt{3} t}{2} \right)} + 15 \cos{\left(\frac{5 \sqrt{3} t}{2} \right)}\right) e^{- \frac{5 t}{2}}}{6} + 10 \frac{d}{d t} \frac{\left(13 \sqrt{3} \sin{\left(\frac{5 \sqrt{3} t}{2} \right)} + 15 \cos{\left(\frac{5 \sqrt{3} t}{2} \right)}\right) e^{- \frac{5 t}{2}}}{60} + 2 \frac{d^{2}}{d t^{2}} \frac{\left(13 \sqrt{3} \sin{\left(\frac{5 \sqrt{3} t}{2} \right)} + 15 \cos{\left(\frac{5 \sqrt{3} t}{2} \right)}\right) e^{- \frac{5 t}{2}}}{60} = 0 \]
    This x(t) satisfies it ...✓ Proved
  3. \[ \left. \frac{\left(13 \sqrt{3} \sin{\left(\frac{5 \sqrt{3} t}{2} \right)} + 15 \cos{\left(\frac{5 \sqrt{3} t}{2} \right)}\right) e^{- \frac{5 t}{2}}}{60} \right|_{\substack{ t=0 }} = \frac{1}{4} \]
    ... starts in the right place ...✓ Proved
  4. \[ \left. \frac{d}{d t} \frac{\left(13 \sqrt{3} \sin{\left(\frac{5 \sqrt{3} t}{2} \right)} + 15 \cos{\left(\frac{5 \sqrt{3} t}{2} \right)}\right) e^{- \frac{5 t}{2}}}{60} \right|_{\substack{ t=0 }} = 1 \]
    ... with the right velocity.✓ Proved
Answer \( x(t) = \frac{\left(13 \sqrt{3} \sin{\left(\frac{5 \sqrt{3} t}{2} \right)} + 15 \cos{\left(\frac{5 \sqrt{3} t}{2} \right)}\right) e^{- \frac{5 t}{2}}}{60} \)

✓ Nihil obstat Lines: 3 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 6.496

medium
A weight of \( \displaystyle 16 \) lb stretches a spring \( \displaystyle \frac{1}{4} \) ft. There is no damping. It is released from rest \( \displaystyle \frac{1}{2} \) ft below equilibrium. Find the position x(t).
  1. \[ \frac{16}{1 \cdot \frac{1}{4}} = 64 \]
    Hooke's law: k = weight / stretch (lb/ft).✓ Proved
  2. \[ \frac{16}{32} = \frac{1}{2} \]
    m = weight / g, with g = 32 ft/s².✓ Proved
  3. The motion satisfies m x″ + c x′ + k x = 0: 1/2x″ + 0x′ + 64x = 0 (x measured downward from equilibrium).
    Reviewed
  4. \[ 32 \cos{\left(8 \sqrt{2} t \right)} + \frac{\frac{d^{2}}{d t^{2}} \frac{\cos{\left(8 \sqrt{2} t \right)}}{2}}{2} = 0 \]
    This x(t) satisfies it ...✓ Proved
  5. \[ \left. \frac{\cos{\left(8 \sqrt{2} t \right)}}{2} \right|_{\substack{ t=0 }} = \frac{1}{2} \]
    ... starts in the right place ...✓ Proved
  6. \[ \left. \frac{d}{d t} \frac{\cos{\left(8 \sqrt{2} t \right)}}{2} \right|_{\substack{ t=0 }} = 0 \]
    ... with the right velocity.✓ Proved
Answer \( x(t) = \frac{\cos{\left(8 \sqrt{2} t \right)}}{2} \)

✓ Nihil obstat Lines: 5 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 6.658

medium
A weight of \( \displaystyle 8 \) lb stretches a spring \( \displaystyle \frac{2}{3} \) ft. There is no damping. It starts \( \displaystyle \frac{1}{2} \) ft below equilibrium with velocity \( \displaystyle -2 \) ft/s (down is positive). Find the position x(t).
  1. \[ \frac{8}{2 \cdot \frac{1}{3}} = 12 \]
    Hooke's law: k = weight / stretch (lb/ft).✓ Proved
  2. \[ \frac{8}{32} = \frac{1}{4} \]
    m = weight / g, with g = 32 ft/s².✓ Proved
  3. The motion satisfies m x″ + c x′ + k x = 0: 1/4x″ + 0x′ + 12x = 0 (x measured downward from equilibrium).
    Reviewed
  4. \[ - 2 \sqrt{3} \sin{\left(4 \sqrt{3} t \right)} + 6 \cos{\left(4 \sqrt{3} t \right)} + \frac{\frac{d^{2}}{d t^{2}} \left(- \frac{\sqrt{3} \sin{\left(4 \sqrt{3} t \right)}}{6} + \frac{\cos{\left(4 \sqrt{3} t \right)}}{2}\right)}{4} = 0 \]
    This x(t) satisfies it ...✓ Proved
  5. \[ \left. \frac{- \sqrt{3} \sin{\left(4 \sqrt{3} t \right)}}{6} + \frac{\cos{\left(4 \sqrt{3} t \right)}}{2} \right|_{\substack{ t=0 }} = \frac{1}{2} \]
    ... starts in the right place ...✓ Proved
  6. \[ \left. \frac{d}{d t} \left(\frac{- \sqrt{3} \sin{\left(4 \sqrt{3} t \right)}}{6} + \frac{\cos{\left(4 \sqrt{3} t \right)}}{2}\right) \right|_{\substack{ t=0 }} = -2 \]
    ... with the right velocity.✓ Proved
Answer \( x(t) = - \frac{\sqrt{3} \sin{\left(4 \sqrt{3} t \right)}}{6} + \frac{\cos{\left(4 \sqrt{3} t \right)}}{2} \)

✓ Nihil obstat Lines: 5 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 6.665

medium
A weight of \( \displaystyle 32 \) lb stretches a spring \( \displaystyle \frac{2}{3} \) ft. There is no damping. It starts \( \displaystyle 1 \) ft below equilibrium with velocity \( \displaystyle 1 \) ft/s (down is positive). Find the position x(t).
  1. \[ \frac{32}{2 \cdot \frac{1}{3}} = 48 \]
    Hooke's law: k = weight / stretch (lb/ft).✓ Proved
  2. \[ \frac{32}{32} = 1 \]
    m = weight / g, with g = 32 ft/s².✓ Proved
  3. The motion satisfies m x″ + c x′ + k x = 0: 1x″ + 0x′ + 48x = 0 (x measured downward from equilibrium).
    Reviewed
  4. \[ 4 \sqrt{3} \sin{\left(4 \sqrt{3} t \right)} + 48 \cos{\left(4 \sqrt{3} t \right)} + \frac{d^{2}}{d t^{2}} \left(\frac{\sqrt{3} \sin{\left(4 \sqrt{3} t \right)}}{12} + \cos{\left(4 \sqrt{3} t \right)}\right) = 0 \]
    This x(t) satisfies it ...✓ Proved
  5. \[ \left. \frac{\sqrt{3} \sin{\left(4 \sqrt{3} t \right)}}{12} + \cos{\left(4 \sqrt{3} t \right)} \right|_{\substack{ t=0 }} = 1 \]
    ... starts in the right place ...✓ Proved
  6. \[ \left. \frac{d}{d t} \left(\frac{\sqrt{3} \sin{\left(4 \sqrt{3} t \right)}}{12} + \cos{\left(4 \sqrt{3} t \right)}\right) \right|_{\substack{ t=0 }} = 1 \]
    ... with the right velocity.✓ Proved
Answer \( x(t) = \frac{\sqrt{3} \sin{\left(4 \sqrt{3} t \right)}}{12} + \cos{\left(4 \sqrt{3} t \right)} \)

✓ Nihil obstat Lines: 5 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 6.633

medium
A \( \displaystyle 5 \) kg mass hangs from a spring with constant \( \displaystyle 20 \) N/m. There is no damping. It is released from rest \( \displaystyle \frac{1}{2} \) m below equilibrium. Find the position x(t).
  1. The motion satisfies m x″ + c x′ + k x = 0: 5x″ + 0x′ + 20x = 0 (x measured downward from equilibrium).
    Reviewed
  2. \[ 10 \cos{\left(2 t \right)} + 5 \frac{d^{2}}{d t^{2}} \frac{\cos{\left(2 t \right)}}{2} = 0 \]
    This x(t) satisfies it ...✓ Proved
  3. \[ \left. \frac{\cos{\left(2 t \right)}}{2} \right|_{\substack{ t=0 }} = \frac{1}{2} \]
    ... starts in the right place ...✓ Proved
  4. \[ \left. \frac{d}{d t} \frac{\cos{\left(2 t \right)}}{2} \right|_{\substack{ t=0 }} = 0 \]
    ... with the right velocity.✓ Proved
Answer \( x(t) = \frac{\cos{\left(2 t \right)}}{2} \)

✓ Nihil obstat Lines: 3 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 6.638

medium
A \( \displaystyle 4 \) kg mass hangs from a spring with constant \( \displaystyle 144 \) N/m. There is no damping. It starts \( \displaystyle \frac{1}{4} \) m below equilibrium with velocity \( \displaystyle 1 \) m/s (down is positive). Find the position x(t).
  1. The motion satisfies m x″ + c x′ + k x = 0: 4x″ + 0x′ + 144x = 0 (x measured downward from equilibrium).
    Reviewed
  2. \[ 24 \sin{\left(6 t \right)} + 36 \cos{\left(6 t \right)} + 4 \frac{d^{2}}{d t^{2}} \left(\frac{\sin{\left(6 t \right)}}{6} + \frac{\cos{\left(6 t \right)}}{4}\right) = 0 \]
    This x(t) satisfies it ...✓ Proved
  3. \[ \left. \frac{\sin{\left(6 t \right)}}{6} + \frac{\cos{\left(6 t \right)}}{4} \right|_{\substack{ t=0 }} = \frac{1}{4} \]
    ... starts in the right place ...✓ Proved
  4. \[ \left. \frac{d}{d t} \left(\frac{\sin{\left(6 t \right)}}{6} + \frac{\cos{\left(6 t \right)}}{4}\right) \right|_{\substack{ t=0 }} = 1 \]
    ... with the right velocity.✓ Proved
Answer \( x(t) = \frac{\sin{\left(6 t \right)}}{6} + \frac{\cos{\left(6 t \right)}}{4} \)

✓ Nihil obstat Lines: 3 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 6.662

medium
A \( \displaystyle 2 \) kg mass hangs from a spring with constant \( \displaystyle 50 \) N/m. There is no damping. It starts \( \displaystyle 1 \) m below equilibrium with velocity \( \displaystyle -2 \) m/s (down is positive). Find the position x(t).
  1. The motion satisfies m x″ + c x′ + k x = 0: 2x″ + 0x′ + 50x = 0 (x measured downward from equilibrium).
    Reviewed
  2. \[ - 20 \sin{\left(5 t \right)} + 50 \cos{\left(5 t \right)} + 2 \frac{d^{2}}{d t^{2}} \left(- \frac{2 \sin{\left(5 t \right)}}{5} + \cos{\left(5 t \right)}\right) = 0 \]
    This x(t) satisfies it ...✓ Proved
  3. \[ \left. - \frac{2 \sin{\left(5 t \right)}}{5} + \cos{\left(5 t \right)} \right|_{\substack{ t=0 }} = 1 \]
    ... starts in the right place ...✓ Proved
  4. \[ \left. \frac{d}{d t} \left(- \frac{2 \sin{\left(5 t \right)}}{5} + \cos{\left(5 t \right)}\right) \right|_{\substack{ t=0 }} = -2 \]
    ... with the right velocity.✓ Proved
Answer \( x(t) = - \frac{2 \sin{\left(5 t \right)}}{5} + \cos{\left(5 t \right)} \)

✓ Nihil obstat Lines: 3 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

More spring-mass systems practice

A weight of \( \displaystyle 4 \) lb stretches a spring \( \displaystyle \frac{1}{2} \) ft. There is no damping. It is released from rest \( \displaystyle \frac{1}{2} \) ft above equilibrium. Find the position x(t).
Problem 6.487medium✓ Every equation proved
A \( \displaystyle 2 \) kg mass hangs from a spring with constant \( \displaystyle 8 \) N/m. There is no damping. It starts \( \displaystyle \frac{1}{3} \) m below equilibrium with velocity \( \displaystyle -2 \) m/s (down is positive). Find the position x(t).
Problem 6.495medium✓ Every equation proved
A \( \displaystyle 1 \) kg mass hangs from a spring with constant \( \displaystyle 9 \) N/m. There is no damping. It starts \( \displaystyle \frac{1}{4} \) m below equilibrium with velocity \( \displaystyle 1 \) m/s (down is positive). Find the position x(t).
Problem 6.498medium✓ Every equation proved
A \( \displaystyle 2 \) kg mass hangs from a spring with constant \( \displaystyle 72 \) N/m. There is no damping. It is released from rest \( \displaystyle \frac{1}{2} \) m above equilibrium. Find the position x(t).
Problem 6.631medium✓ Every equation proved
A weight of \( \displaystyle 2 \) lb stretches a spring \( \displaystyle \frac{1}{3} \) ft. There is no damping. It is released from rest \( \displaystyle \frac{1}{3} \) ft below equilibrium. Find the position x(t).
Problem 6.632medium✓ Nihil obstat
A weight of \( \displaystyle 64 \) lb stretches a spring \( \displaystyle 1 \) ft. There is no damping. It starts \( \displaystyle \frac{1}{2} \) ft above equilibrium with velocity \( \displaystyle 1 \) ft/s (down is positive). Find the position x(t).
Problem 6.635medium✓ Every equation proved
A weight of \( \displaystyle 64 \) lb stretches a spring \( \displaystyle \frac{1}{3} \) ft. There is no damping. It is released from rest \( \displaystyle \frac{1}{2} \) ft above equilibrium. Find the position x(t).
Problem 6.637medium✓ Every equation proved
A \( \displaystyle 4 \) kg mass hangs from a spring with constant \( \displaystyle 16 \) N/m. There is no damping. It is released from rest \( \displaystyle \frac{1}{3} \) m below equilibrium. Find the position x(t).
Problem 6.639medium✓ Every equation proved
A \( \displaystyle 1 \) kg mass hangs from a spring with constant \( \displaystyle 4 \) N/m. There is no damping. It is released from rest \( \displaystyle \frac{1}{2} \) m above equilibrium. Find the position x(t).
Problem 6.640medium✓ Every equation proved
A \( \displaystyle 1 \) kg mass hangs from a spring with constant \( \displaystyle 4 \) N/m. There is no damping. It is released from rest \( \displaystyle \frac{1}{4} \) m below equilibrium. Find the position x(t).
Problem 6.642medium✓ Every equation proved
A weight of \( \displaystyle 8 \) lb stretches a spring \( \displaystyle \frac{1}{2} \) ft. There is no damping. It is released from rest \( \displaystyle \frac{1}{3} \) ft below equilibrium. Find the position x(t).
Problem 6.643medium✓ Nihil obstat
A weight of \( \displaystyle 16 \) lb stretches a spring \( \displaystyle 2 \) ft. There is no damping. It is released from rest \( \displaystyle \frac{1}{2} \) ft below equilibrium. Find the position x(t).
Problem 6.645medium✓ Nihil obstat
A \( \displaystyle 4 \) kg mass hangs from a spring with constant \( \displaystyle 36 \) N/m. There is no damping. It is released from rest \( \displaystyle \frac{1}{3} \) m below equilibrium. Find the position x(t).
Problem 6.661medium✓ Nihil obstat
A weight of \( \displaystyle 64 \) lb stretches a spring \( \displaystyle 1 \) ft. There is no damping. It starts \( \displaystyle \frac{1}{3} \) ft below equilibrium with velocity \( \displaystyle 1 \) ft/s (down is positive). Find the position x(t).
Problem 6.667medium✓ Nihil obstat
A \( \displaystyle 5 \) kg mass hangs from a spring with constant \( \displaystyle 5 \) N/m. There is no damping. It is released from rest \( \displaystyle \frac{1}{2} \) m below equilibrium. Find the position x(t).
Problem 6.719medium✓ Every equation proved
A \( \displaystyle 4 \) kg mass hangs from a spring with constant \( \displaystyle 100 \) N/m. There is no damping. It starts \( \displaystyle \frac{1}{3} \) m below equilibrium with velocity \( \displaystyle 1 \) m/s (down is positive). Find the position x(t).
Problem 6.720medium✓ Every equation proved
A \( \displaystyle 5 \) kg mass hangs from a spring with constant \( \displaystyle 180 \) N/m. There is no damping. It is released from rest \( \displaystyle 1 \) m below equilibrium. Find the position x(t).
Problem 6.722medium✓ Nihil obstat
A weight of \( \displaystyle 32 \) lb stretches a spring \( \displaystyle 2 \) ft. There is no damping. It starts \( \displaystyle \frac{1}{2} \) ft below equilibrium with velocity \( \displaystyle 1 \) ft/s (down is positive). Find the position x(t).
Problem 6.723medium✓ Nihil obstat
A weight of \( \displaystyle 16 \) lb stretches a spring \( \displaystyle 2 \) ft. There is no damping. It starts \( \displaystyle \frac{1}{4} \) ft below equilibrium with velocity \( \displaystyle -2 \) ft/s (down is positive). Find the position x(t).
Problem 6.724medium✓ Nihil obstat
A weight of \( \displaystyle 32 \) lb stretches a spring \( \displaystyle 2 \) ft. There is no damping. It starts \( \displaystyle \frac{1}{4} \) ft below equilibrium with velocity \( \displaystyle -2 \) ft/s (down is positive). Find the position x(t).
Problem 6.725medium✓ Nihil obstat
A \( \displaystyle 2 \) kg mass hangs from a spring with constant \( \displaystyle 2 \) N/m. There is no damping. It is released from rest \( \displaystyle 1 \) m below equilibrium. Find the position x(t).
Problem 6.726medium✓ Every equation proved
A \( \displaystyle 4 \) kg mass hangs from a spring with constant \( \displaystyle 4 \) N/m. There is no damping. It starts \( \displaystyle \frac{1}{4} \) m below equilibrium with velocity \( \displaystyle 1 \) m/s (down is positive). Find the position x(t).
Problem 6.728medium✓ Nihil obstat
A \( \displaystyle 5 \) kg mass hangs from a spring with constant \( \displaystyle 45 \) N/m. There is no damping. It is released from rest \( \displaystyle 1 \) m below equilibrium. Find the position x(t).
Problem 6.809medium✓ Nihil obstat
A \( \displaystyle 1 \) kg mass hangs from a spring with constant \( \displaystyle 16 \) N/m. There is no damping. It starts \( \displaystyle 1 \) m below equilibrium with velocity \( \displaystyle -2 \) m/s (down is positive). Find the position x(t).
Problem 6.810medium✓ Nihil obstat
A weight of \( \displaystyle 16 \) lb stretches a spring \( \displaystyle \frac{1}{2} \) ft. There is no damping. It starts \( \displaystyle \frac{1}{2} \) ft below equilibrium with velocity \( \displaystyle 1 \) ft/s (down is positive). Find the position x(t).
Problem 6.811medium✓ Nihil obstat
A weight of \( \displaystyle 32 \) lb stretches a spring \( \displaystyle 1 \) ft. There is no damping. It is released from rest \( \displaystyle 1 \) ft below equilibrium. Find the position x(t).
Problem 6.812medium✓ Nihil obstat
A weight of \( \displaystyle 8 \) lb stretches a spring \( \displaystyle 1 \) ft. There is no damping. It starts \( \displaystyle \frac{1}{2} \) ft below equilibrium with velocity \( \displaystyle -2 \) ft/s (down is positive). Find the position x(t).
Problem 6.813medium✓ Nihil obstat
A \( \displaystyle 4 \) kg mass hangs from a spring with constant \( \displaystyle 64 \) N/m. There is no damping. It is released from rest \( \displaystyle 1 \) m below equilibrium. Find the position x(t).
Problem 6.814medium✓ Nihil obstat
A weight of \( \displaystyle 32 \) lb stretches a spring \( \displaystyle \frac{2}{3} \) ft. There is no damping. It is released from rest \( \displaystyle \frac{1}{3} \) ft below equilibrium. Find the position x(t).
Problem 6.815medium✓ Nihil obstat
A weight of \( \displaystyle 4 \) lb stretches a spring \( \displaystyle 2 \) ft. There is no damping. It starts \( \displaystyle \frac{1}{2} \) ft below equilibrium with velocity \( \displaystyle 1 \) ft/s (down is positive). Find the position x(t).
Problem 6.816medium✓ Nihil obstat
A weight of \( \displaystyle 4 \) lb stretches a spring \( \displaystyle \frac{1}{2} \) ft. There is no damping. It is released from rest \( \displaystyle \frac{1}{4} \) ft below equilibrium. Find the position x(t).
Problem 6.818medium✓ Every equation proved
A \( \displaystyle 2 \) kg mass hangs from a spring with constant \( \displaystyle 2 \) N/m. A damping force of \( \displaystyle 2 \) times the velocity acts on it. It is released from rest \( \displaystyle \frac{1}{3} \) m below equilibrium. Find the position x(t).
Problem 6.489hard✓ Nihil obstat
A \( \displaystyle 4 \) kg mass hangs from a spring with constant \( \displaystyle 64 \) N/m. A damping force of \( \displaystyle 40 \) times the velocity acts on it. It starts \( \displaystyle \frac{1}{4} \) m below equilibrium with velocity \( \displaystyle -2 \) m/s (down is positive). Find the position x(t).
Problem 6.492hard✓ Every equation proved
A \( \displaystyle 4 \) kg mass hangs from a spring with constant \( \displaystyle 4 \) N/m. A damping force of \( \displaystyle 10 \) times the velocity acts on it. It is released from rest \( \displaystyle \frac{1}{2} \) m below equilibrium. Find the position x(t).
Problem 6.493hard✓ Nihil obstat
A weight of \( \displaystyle 2 \) lb stretches a spring \( \displaystyle \frac{1}{2} \) ft. A damping force of \( \displaystyle \frac{1}{2} \) times the velocity acts on it. It starts \( \displaystyle \frac{1}{3} \) ft below equilibrium with velocity \( \displaystyle 1 \) ft/s (down is positive). Find the position x(t).
Problem 6.494hard✓ Nihil obstat
A \( \displaystyle 4 \) kg mass hangs from a spring with constant \( \displaystyle 100 \) N/m. A damping force of \( \displaystyle 50 \) times the velocity acts on it. It is released from rest \( \displaystyle \frac{1}{2} \) m above equilibrium. Find the position x(t).
Problem 6.497hard✓ Every equation proved
A \( \displaystyle 2 \) kg mass hangs from a spring with constant \( \displaystyle 18 \) N/m. A damping force of \( \displaystyle 6 \) times the velocity acts on it. It starts \( \displaystyle 1 \) m below equilibrium with velocity \( \displaystyle -2 \) m/s (down is positive). Find the position x(t).
Problem 6.499hard✓ Nihil obstat
A \( \displaystyle 5 \) kg mass hangs from a spring with constant \( \displaystyle 20 \) N/m. A damping force of \( \displaystyle 20 \) times the velocity acts on it. It is released from rest \( \displaystyle \frac{1}{3} \) m below equilibrium. Find the position x(t).
Problem 6.500hard✓ Every equation proved
A \( \displaystyle 2 \) kg mass hangs from a spring with constant \( \displaystyle 50 \) N/m. A damping force of \( \displaystyle 10 \) times the velocity acts on it. It is released from rest \( \displaystyle \frac{1}{2} \) m above equilibrium. Find the position x(t).
Problem 6.634hard✓ Every equation proved
A \( \displaystyle 5 \) kg mass hangs from a spring with constant \( \displaystyle 125 \) N/m. A damping force of \( \displaystyle \frac{125}{2} \) times the velocity acts on it. It starts \( \displaystyle \frac{1}{2} \) m below equilibrium with velocity \( \displaystyle 1 \) m/s (down is positive). Find the position x(t).
Problem 6.636hard✓ Nihil obstat
A \( \displaystyle 2 \) kg mass hangs from a spring with constant \( \displaystyle 2 \) N/m. A damping force of \( \displaystyle \frac{12}{5} \) times the velocity acts on it. It is released from rest \( \displaystyle \frac{1}{3} \) m below equilibrium. Find the position x(t).
Problem 6.641hard✓ Nihil obstat
A \( \displaystyle 2 \) kg mass hangs from a spring with constant \( \displaystyle 2 \) N/m. A damping force of \( \displaystyle 4 \) times the velocity acts on it. It starts \( \displaystyle 1 \) m below equilibrium with velocity \( \displaystyle -2 \) m/s (down is positive). Find the position x(t).
Problem 6.644hard✓ Every equation proved
A \( \displaystyle 1 \) kg mass hangs from a spring with constant \( \displaystyle 9 \) N/m. A damping force of \( \displaystyle \frac{15}{2} \) times the velocity acts on it. It starts \( \displaystyle \frac{1}{2} \) m above equilibrium with velocity \( \displaystyle 1 \) m/s (down is positive). Find the position x(t).
Problem 6.659hard✓ Every equation proved
A \( \displaystyle 2 \) kg mass hangs from a spring with constant \( \displaystyle 50 \) N/m. A damping force of \( \displaystyle 25 \) times the velocity acts on it. It starts \( \displaystyle \frac{1}{2} \) m above equilibrium with velocity \( \displaystyle -2 \) m/s (down is positive). Find the position x(t).
Problem 6.660hard✓ Every equation proved
A \( \displaystyle 2 \) kg mass hangs from a spring with constant \( \displaystyle 2 \) N/m. A damping force of \( \displaystyle 5 \) times the velocity acts on it. It starts \( \displaystyle \frac{1}{3} \) m below equilibrium with velocity \( \displaystyle -2 \) m/s (down is positive). Find the position x(t).
Problem 6.663hard✓ Nihil obstat
A weight of \( \displaystyle 8 \) lb stretches a spring \( \displaystyle \frac{1}{2} \) ft. A damping force of \( \displaystyle 4 \) times the velocity acts on it. It starts \( \displaystyle \frac{1}{2} \) ft above equilibrium with velocity \( \displaystyle 1 \) ft/s (down is positive). Find the position x(t).
Problem 6.666hard✓ Every equation proved
A \( \displaystyle 4 \) kg mass hangs from a spring with constant \( \displaystyle 144 \) N/m. A damping force of \( \displaystyle 60 \) times the velocity acts on it. It is released from rest \( \displaystyle \frac{1}{2} \) m above equilibrium. Find the position x(t).
Problem 6.721hard✓ Every equation proved
A weight of \( \displaystyle 32 \) lb stretches a spring \( \displaystyle 2 \) ft. A damping force of \( \displaystyle 4 \) times the velocity acts on it. It starts \( \displaystyle \frac{1}{4} \) ft below equilibrium with velocity \( \displaystyle -2 \) ft/s (down is positive). Find the position x(t).
Problem 6.727hard✓ Every equation proved
A \( \displaystyle 5 \) kg mass hangs from a spring with constant \( \displaystyle 80 \) N/m. A damping force of \( \displaystyle 40 \) times the velocity acts on it. It is released from rest \( \displaystyle 1 \) m below equilibrium. Find the position x(t).
Problem 6.817hard✓ Nihil obstat