Using Spring Equations - AP Physics C: Electricity and Magnetism

Card 0 of 132

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 $\small m$ ?

Tap to see back →

Relevant equations:

$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.

$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.

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

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

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

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?

Tap to see back →

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

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

Rearrange to solve for the velocity:

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:

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

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

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?

Tap to see back →

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:

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?

Tap to see back →

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.

Use our given values to solve:

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?

Tap to see back →

When a mass is in free fall, its potential energy is increasing. In this instance, all of that energy must be counteracted by the bungee cord, which we can treat like a spring , for the person to come to a stop. This means that we can set these two equations as equal to one another:

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)}$

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?

Tap to see back →

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:

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

A mass on the end of a spring oscillates back and forth. The period of an oscillating spring is measured to be 2.27 seconds. The mass is measured to be 0.300 kg. What is the spring's spring constant?

Tap to see back →

To solve this problem, we can use the following equation:

$\small T=2\pi$\sqrt{\frac{m}{k}$}$

Here, is the period of an oscillation, is the mass of the oscillating thing in kg, and is the spring constant in .

Rearranging the equation to solve for the spring constant, , we get:

$k\cdot $=\frac{m}{(\frac{T}${2\pi})^2$}$

Plugging in the two known values, we get:

$k\cdot $=\frac{0.300kg}{(\frac{2.27s}${2\pi})^2$}$

Solving for , we get .

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?

Tap to see back →

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.

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?

Tap to see back →

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

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

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?

Tap to see back →

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.

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

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 ?

Tap to see back →

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}$$

A block on a frictionless table is connected to a horizontal spring with constant . If the block is released from rest when the spring is stretched a distance of ,

what is its speed when the spring is compressed a distance of ?

Tap to see back →

Assuming this is a frictionless table, we don't have to take the work done by friction into account.

This is a conservation of energy problem. In this problem we have to realize that the potential energy of the spring at is equal to the kinetic energy of the spring at + the potential energy of the spring at .

This can be shown in the following equation:

$\frac{1}{2}$kx_$7^2$ = $$\frac{1}{2}$mv^2$ + $\frac{1}{2}$kx_$3^2$

If we solve for , we get the following equation:

$v = $\sqrt{2\cdot\frac{\frac{1}{2}$kx_$7^2$-$\frac{1}{2}$kx_$3^2$}{m}}$

Remember to convert the distances given in centimeters to meters.

If we plug in all the variables, we get

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 $\small m$ ?

Tap to see back →

Relevant equations:

$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.

$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.

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

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

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

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?

Tap to see back →

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

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

Rearrange to solve for the velocity:

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:

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

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

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?

Tap to see back →

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:

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?

Tap to see back →

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.

Use our given values to solve:

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?

Tap to see back →

When a mass is in free fall, its potential energy is increasing. In this instance, all of that energy must be counteracted by the bungee cord, which we can treat like a spring , for the person to come to a stop. This means that we can set these two equations as equal to one another:

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)}$

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?

Tap to see back →

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:

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

A mass on the end of a spring oscillates back and forth. The period of an oscillating spring is measured to be 2.27 seconds. The mass is measured to be 0.300 kg. What is the spring's spring constant?

Tap to see back →

To solve this problem, we can use the following equation:

$\small T=2\pi$\sqrt{\frac{m}{k}$}$

Here, is the period of an oscillation, is the mass of the oscillating thing in kg, and is the spring constant in .

Rearranging the equation to solve for the spring constant, , we get:

$k\cdot $=\frac{m}{(\frac{T}${2\pi})^2$}$

Plugging in the two known values, we get:

$k\cdot $=\frac{0.300kg}{(\frac{2.27s}${2\pi})^2$}$

Solving for , we get .

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?

Tap to see back →

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.