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AP Physics C: Mechanics : Using Spring Equations

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All AP Physics C: Mechanics Resources

2 Diagnostic Tests 92 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Using Spring Equations

A mass $\small m$ is attached to a spring of force constant $\small k$ . The mass rests on a frictionless surface and oscillates horizontally, with oscillations of amplitude $\small A$ . What is the maximum velocity of this mass in terms of $\small k$ , $\small A$ , and $\small m$ ?

Possible Answers:

$\sqrt{\frac{kA^2}{m}}$

$\sqrt{\frac{2kA}{m}}$

$\sqrt{\frac{kA}{m}}$

$\sqrt{\frac{kA}{2m}}$

$\sqrt{\frac{kA^2}{2m}}$

Correct answer:

$\sqrt{\frac{kA^2}{m}}$

Explanation:

Relevant equations:

K_i+U_i=K_f+U_f

$U_{sp} = \frac{1}{2}kx^2$

$K = \frac{1}{2}mv^2$

Write expressions for the initial kinetic and potential energies, if the spring is initially stretched to the maximum amplitude before being released.

K_i = 0

$U_i = U_{sp, max}=\frac{1}{2}kA^2$

Write expressions for the final kinetic and potential energies when the spring crosses the equilibrium point.

$K_f = \frac{1}{2}mv_{max}^2$

$U_f = U_{sp,min}=\frac{1}{2}k(0)^2 = 0$

Use conservation of energy to equate the initial and final energy sums.

$0+\frac{1}{2}kA^2=\frac{1}{2}mv_{max}^2 +0$

Solve the equation to isolate $\small v_{max}$ .

$\frac{1}{2}kA^2=\frac{1}{2}mv_{max}^2$

$v_{max}^2=\frac{\frac{1}{2}kA^2}{\frac{1}{2}m}$

$v_{max}=\sqrt{\frac{kA^2}{m}}$

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Example Question #2 : Using Spring Equations

A ball is attached to a spring on a frictionless, horizontal plane. If the spring constant is $12\frac{N}{m}$ and the mass of the ball three kilograms, at what angular frequency will the system oscillate?

Possible Answers:

$2\frac{rad}{s}$

$3\frac{rad}{s}$

$4\frac{rad}{s}$

$5\frac{rad}{s}$

$1\frac{rad}{s}$

Correct answer:

$2\frac{rad}{s}$

Explanation:

The units for angular frequency, $\omega$ , are radians per second.

We need to derive the equation for angular frequency using conservation of energy.

$KE=PE_{spring}$

$\frac{1}{2}mv^2=\frac{1}{2}kx^2$

Rearrange to solve for the velocity:

$v=\sqrt{\frac{kx^2}{m}}$

The velocity is also the product of angular frequency and the distance of the oscillation:

$v=\omega x$

Use this equation to derive the equation for angular frequency:

$\omega=\frac{v}{x}=\frac{\sqrt{\frac{kx^2}{m}}}{x}=\sqrt{\frac{k}{m}}$

Finally, use our given mass and spring constant to solve:

$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{12}{3}} = 2\ \frac{rad}{s}$

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Example Question #1 : Using Spring Equations

A 1kg ball is attached to a massless spring on a frictionless, horizontal plane. If at its equilibrium position, the ball is moving at $4\frac{m}{s}$ , how much total energy is in the system?

Possible Answers:

16J

14J

10J

12J

Correct answer:

Explanation:

At the equilibrium position, the spring does not contribute any potential energy. It is neither stretched, nor compressed.

$PE_{spring}=\frac{1}{2}kx^2\rightarrow \frac{1}{2}k(0)=0$

All of the energy in the system is kinetic energy, resulting from the given velocity:

$KE=\frac{1}{2}mv^2$

$KE=\frac{1}{2}(1kg)(4\frac{m}{s})^2=8J$

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Example Question #3 : Using Spring Equations

A 500g ball is attached to a massless spring on a frictionless, horizontal plane. If at its equilibrium position, the ball is moving at $4\frac{m}{s}$ , and the spring constant is $0.125\frac{N}{m}$ , what is the maximum displacement of the ball from its equilibrium position?

Possible Answers:

10m

0.5m

Correct answer:

Explanation:

To solve this question, we will need to use conservation of energy. With no displacement, the ball has a velocity of $4\frac{m}{s}$ and zero displacement. At its maximum displacement, the velocity will be zero and all the energy will be converted to spring potential energy.

KE_i=mv_i^2

$PE_{f}=\frac{1}{2}kx_f^2$

KE_i=PE_f

mv_i^2=kx_f^2

Use our given values to solve:

$(0.5kg)(4\frac{m}{s})^2=(0.125\frac{N}{m})x^2$

$x=\sqrt{\frac{(0.5kg)(4\frac{m}{s})^2}{0.125\frac{N}{m}}}=8m$

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Example Question #2 : Using Spring Equations

A bungee-jumping company operates on a bridge 200 m above the ground. They use bungee cords that are 100 m when they are unstretched; these bungee cords have a spring constant of $25 \frac{N}{m}$ . What could the mass of their largest customer be before hitting the ground becomes an issue?

$g=9.81\frac{m}{s^2}$

Possible Answers:

100kg

127kg

135kg

150kg

90kg

Correct answer:

127kg

Explanation:

When a mass is in free fall, its potential energy (K=mgh) is increasing. In this instance, all of that energy must be counteracted by the bungee cord, which we can treat like a spring $(K=\frac{1}{2}kx^{2})$ , for the person to come to a stop. This means that we can set these two equations as equal to one another:

$mgh=\frac{1}{2}kx^{2}$

We can then rearrange this equation to solve for , the mass at which the bungee cord can be stretched as far as the ground:

$m=\frac{k\cdot x^2}{2\cdot g\cdot h}$

Substituting in the known values, we can solve for :

$m=\frac{k(total\:height - length\:of\:cord)^{2}}{2\cdot g\cdot (length\:of\:cord)}$

$m=\frac{25\frac{N}{m}(200m-100m)^2}{2\cdot 9.81\frac{m}{s^2}\cdot100m}$

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Example Question #1 : Using Spring Equations

An 85 kg stuntman stands on a spring-loaded platform for a movie scene. The spring constant for the platform is $1200 \frac{N}{m}$ . How far should the platform be compressed to launch the stuntman 7.0 m in the air?

Possible Answers:

6.2m

2.3m

4.7m

3.1m

1.6m

Correct answer:

3.1m

Explanation:

All the energy that is stored up by the compressed spring needs to turn into gravitational potential energy for the stuntman to come to a rest 7.0 m in the air. Gravitational potential energy is modeled by the equation $\small K=mgh$ and the energy in the spring system is modeled by the equation $\small K=\frac{1}{2}kx^{2}$ . We can thus set these equations equal to one another:

$mgh=\frac{1}{2}kx^2$

We then rearrange the resulting equation for the compressed distance, which is :

$x=\sqrt{\frac{2mgh}{k}}$

Substituting in the given values, we can solve for :

$x=\sqrt{\frac{2(85kg)(9.81\frac{m}{s^2})(7.0 m)}{1200\frac{N}{m}}}$

$x=\sqrt{9.72825m}= 3.11901426736m\approx3.1m$

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Example Question #5 : Using Spring Equations

A spring with spring constant $15\frac{N}{cm}$ is attached to a 3.0 kg mass. The mass is then displaced 5 cm and released. How long will it take for the mass to travel to a point that's 3.0 cm past the spring’s equilibrium point during its first oscillation?

Possible Answers:

0.42s

1.3s

0.92s

1.0s

2.2s

Correct answer:

1.0s

Explanation:

We start with $\small x=Acos(\sigma t)$ , where is the point the mass is traveling to, is the maximum displacement, $\small \sigma$ is $\small \sqrt{\frac{k}{m}}$ , and is the time it takes to travel the distance.

Rearranging this equation for gives

$\small t=\sqrt{\tfrac{m}{k}}\:(cos(\frac{x}{A}))^{-1}$

Here, is because the point we're looking for is on the other side of the equilibrium point.

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Example Question #2 : Using Spring Equations

A 10 kg mass is attached to a vertical spring that is hanging from a ceiling. The spring's constant is $50\frac{N}{m}$ . How much will gravity stretch the spring?

$g=10\frac{m}{s^2}$

Possible Answers:

0.5m

0.2m

Correct answer:

Explanation:

We compare the force from Hooke's Law to the force from gravity and solve for displacement:

F=mg=-kx

Rearranging this equation to solve for , we get

$-x=\frac{mg}{k}$

Plugging in the given values, we get

$-x=\frac{10kg\cdot 10\frac{m}{s^{2}}}{50\frac{N}{m}}$

Solving for , we get

-x=2m

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Example Question #1 : Using Spring Equations

A 3 kg mass is attached to a spring that is attached to a wall. The mass is pulled 10 cm away from the spring's equilibrium point and released. If the spring has a constant of $27\frac{N}{m}$ , what is the maximum velocity the mass will reach?

Possible Answers:

$3.0\frac{m}{s}$

$0.3\frac{m}{s}$

$0.9\frac{m}{s}$

$9.0\frac{m}{s}$

$1.5\frac{m}{s}$

Correct answer:

$0.3\frac{m}{s}$

Explanation:

When the spring is at its maximum compression, its potential energy is maximized and its kinetic energy is 0. When the mass is at the spring's equilibrium point, all of the potential energy is converted into kinetic energy, so kinetic energy is maximized (and equal to the maximum potential energy) and potential energy is 0. We set these two equations equal to each other and solve for velocity.

$K_{max}=U_{max}$

$\frac{1}{2}mv_{max}^{2}=\frac{1}{2}kx_{max}^{2}$

$v_{max}=x_{max}\sqrt{\frac{k}{m}}$

$v_{max}=0.1m\sqrt{\frac{27\frac{N}{m}}{3kg}}$

$v_{max}=0.3\frac{m}{s}$

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Example Question #6 : Using Spring Equations

All angles in this problem are expressed in radians.

An object oscillates horizontally, and its displacement from equilibrium can be found using the equation:

$x=(2.00m)cos(\pi t+\frac{\pi }{2})$

where is in seconds. What is the velocity of the object at t=3.00s ?

Possible Answers:

$2.00\frac{m}{s}$

$-2\pi \frac{m}{s}$

$2\pi \frac{m}{s}$

$0.00\frac{m}{s}$

$\pi\frac{m}{s}$

Correct answer:

$2\pi \frac{m}{s}$

Explanation:

To find the equation for velocity, we take the first derivative of the position function. Don't forget the chain rule for the inside of the cosine!

$v=\frac{d(x)}{dt}$

$v=(-2.00\pi \frac{m}{s})sin(\pi t+\frac{\pi }{2})$

$v=(-2.00\pi \frac{m}{s})sin(\pi (2.00s))+\frac{\pi }{2})$

$v=2\pi \frac{m}{s}$

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All AP Physics C: Mechanics Resources

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AP Physics C: Mechanics : Using Spring Equations

Study concepts, example questions & explanations for AP Physics C: Mechanics

All AP Physics C: Mechanics Resources

Example Questions

Example Question #1 : Using Spring Equations

Example Question #2 : Using Spring Equations

Example Question #1 : Using Spring Equations

Example Question #3 : Using Spring Equations

Example Question #2 : Using Spring Equations

Example Question #1 : Using Spring Equations

Example Question #5 : Using Spring Equations

Example Question #2 : Using Spring Equations

Example Question #1 : Using Spring Equations

Example Question #6 : Using Spring Equations

All AP Physics C: Mechanics Resources

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