AP Physics 1 - Springs | Practice Hub

A object is undergoing SHM with amplitude of . If the spring constant is $250 \frac{N}{m}$ , calculate the maximum speed of the object.

$1.58 \frac{m}{s}$

$12.5 \frac{m}{s}$

$0.80 \frac{m}{s}$

$6.25 \frac{m}{s}$

$3.16 \frac{m}{s}$

Explanation

To solve this problem, we must use our knowledge of the relationship between angular frequency, the spring constant, and the mass.

$\omega = \sqrt{\frac{k}{m}}=\sqrt{\frac{250\frac{N}{m}}{4 kg}}=7.9 \frac{rad}{s^2}$

We can then use this, and the derived maximum speed of the object in terms of the amplitude and angular frequency, to find the answer.

$v_{max}=A\omega=0.2m\left( 7.9\frac{rad}{s}\right )=1.58 \frac{m}{s}$

A horizontal spring with spring constant $100\frac{N}{m}$ is attached to a wall and a mass of . The mass can slide without friction on a frictionless surface.

Determine the frequency of motion of the system if the system is stretched by .

$\omega=\frac{1.41}{s}$

$\omega=\frac{4.55}{s}$

$\omega=\frac{2.00}{s}$

$\omega=\frac{3.33}{s}$

$\omega=\frac{.707}{s}$

Explanation

Use the equation:

$\omega=\sqrt{\frac{k}{m}}$ for a mass on a horizontal spring (so no gravity in the direction of motion)

The stretch length will have no effect.

Plug in values:

$\omega=\sqrt{\frac{100}{50}}$

$\omega=\frac{1.41}{s}$

A spring with a spring constant of $50 \frac{N}{m}$ is compressed past its point of equilibrium. If internal friction results in an average power loss of , how long does the spring oscillate after being released until it comes to rest?

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

Explanation

We can calculate the initial potential energy of the spring from what we are given in the problem statement:

$U = \frac{1}{2}kx^2 = \frac{1}{2}(50\frac{N}{m})(3m)^2 = 225J$

Assuming the only energy loss of the spring is through internal friction, we can write:

$P=\frac{W}{t}=\frac{\Delta U}{t}\rightarrow \Delta U=P\cdot t$

$225J = P\cdot t = (5W)t$

A mass is hung on a vertical spring which extends the spring by 2 meters. What is the spring constant of the spring in $\frac{N}{m}$ ?

$15 \frac{N}{m}$

$20\frac{N}{m}$

$12\frac{N}{m}$

None of these

$30 \frac{N}{m}$

Explanation

The spring constant is equal to $k=\frac{mg}{x}$ .

$k=\frac{3kg*10\frac{m}{s^2}}{2m}=\boldsymbol{15\frac{N}{m}}$

A block falls off of a table and collides with a spring on the floor with velocity $10\frac{m}{s}$ . The spring has a spring constant of $100\frac{N}{m}$ . How much is the spring compressed when the velocity of the block is ?

Explanation

This problem can be solved by the conservation of energy. The block initially has kinetic energy equal to:

$\frac{1}{2}mv^2 = \frac{1}{2}\cdot 5kg\cdot (10\frac{m}{s})^2 = 250 J$

When the velocity of the block is zero, all the kinetic energy has been transferred to potential spring energy.

$250 J = \frac{1}{2}kx^2$

$250 J = \frac{1}{2}\cdot 100 \frac{N}{m} \cdot x^2$

$x = \sqrt{5}=2.24m$

Given that a spring is held above the ground and an object of mass tied to the spring is displaced below the equilibrium position, determine the spring constant .

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

$k=20 \frac{kg}{s^2}$

$k=15 \frac{kg}{s^2}$

$k=60 \frac{kg}{s^2}$

$k=3 \frac{kg}{s^2}$

Explanation

Since there are two forces acting on the object, and it isn't moving, there is an equality of the two forces. This means that:

$F_{spring}=F_{gravity}$

Equivalently:

Where is the spring constant, is the displacement of the spring, is the equilibrium position, is mass of the object and is the gravitational constant given as $10\frac{m}{s^2}$

Since we know that the mass on the spring spring is displaced , the mass is , and we know the gravitational constant we plug in and solve for the spring constant.

$-k*(-3m)=6kg*10\frac{m}{s^2}$

$k=20 \frac{kg}{s^2}$

A homogenous mass of 0.25kg is fixed to a 0.5kg Hookean spring. When the mass/spring system is stretched 1cm from the equilibrium, it takes 3N of force to hold the mass in place. If the displacement from equilibrium is doubled, the force necessary to keep the system in place will __________.

increase by a factor of 2

remain the same

increase by a factor of 4

increase by a factor of $\sqrt{2}$

Explanation

Since the spring is Hookean, the relationship between the force and displacement from equilibrium of the mass can be expressed by Hooke's Law:

Since this equation is linear, the force and displacement are directly proportional. Thus, when the displacement doubles, the force doubles.

A spring with constant $k = 50\ \frac{\textup{N}}{\textup{m}}$ is hanging from a ceiling. A block of mass $4\textup{ kg}$ is attached and the spring is compressed $0.4\textup{ m}$ from equilibrium. The block is then released from rest. What is the velocity of the block as it passes through equilibrium?

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

$3.2\ \frac{\textup{m}}{\textup{s}}$

$2.4\ \frac{\textup{m}}{\textup{s}}$

$1.3\ \frac{\textup{m}}{\textup{s}}$

$4.6\ \frac{\textup{m}}{\textup{s}}$

$5.1\ \frac{\textup{m}}{\textup{s}}$

Explanation

We can use the expression for conservation of energy to solve this problem:

We can eliminate initial kinetic energy (block initially at rest) and final potential energy (block at equilibrium of spring) to get:

Substituting in expressions for each of these, we get:

$mgh_i +\frac{1}{2}kx_i^2=\frac{1}{2}mv_f^2$

Where initial height is simply the displacement of the spring:

$mgx_i +\frac{1}{2}kx_i^2=\frac{1}{2}mv_f^2$

Multiplying both sides of the expression by 2 and rearranging for final velocity, we get:

$v_f = \sqrt{\frac{2mgx_i+kx_i^2}{m}}$

Plugging in values for each of these variables, we get:

$v_ f = \sqrt{\frac{2(4kg)\left ( 10\frac{m}{s^2}\right )(0.4m)+\left ( 50\frac{N}{m}\right )(0.4m)^2}{4kg}}$

$v_f = 3.2\frac{m}{s}$

A horizontal spring with a spring constant of $20 \frac{N}{m}$ sits on a table and has a mass of attached to one end. The coefficient of kinetic friction between the mass and table is . If the spring is stretched to a distance of past its point of equilibrium and released, how many times does the mass pass through the point of equilibrium before coming to rest?

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

Explanation

From the problem statement, we can calculate how much potential energy is initially stored in the spring. Since we are neglecting air resistance, we can say that all of this energy is lost through friction as the system comes to a stop.

Calculating initial potential energy:

$U = \frac{1}{2}kx^2 = \frac{1}{2}(20\frac{N}{m})(2m)^2 = 40 J$

Then we can write an expression for the force of friction on the mass:

$F_f = \mu F_N = (0.20)(1kg)(10\frac{m}{s^2}) = 2N$

We can then write an expression of the work done by friction:

$W = F_f\cdot d = (2N)d$

Setting this equal to the initial potential energy:

Therefore, the mass travels a total distance of 20 meters before coming to rest. We know that the distance between maximum compression and maximum extension of the spring is 4 meters. Therefore, we can say that the mass travels from one maximum displacement to the other five times, passing through the equilibrium once each time. The mass passes through the equilibrium point 5 times.

In the lab, a student has created an oscillator by hanging a weight from a spring. The student releases the oscillator from rest and uses a sensor and computer to find the equation of motion for the oscillator: $x(t)=0.13m;cos(6.3\frac{radians}{s}t+0.14:radians)$

The student then pulls the weight down twice as far and releases it from rest. What would the new equation of motion be for the oscillator?

$x(t)=0.26m;cos(6.3\frac{radians}{s}t+0.14:radians)$

$x(t)=0.13m;cos(12.6\frac{radians}{s}t+0.14:radians)$

$x(t)=0.13m;cos(6.3\frac{radians}{s}t+0.28:radians)$

$x(t)=0.13m;cos(8.9\frac{radians}{s}t+0.14:radians)$

$x(t)=0.18m;cos(6.3\frac{radians}{s}t+0.14:radians)$

Explanation

Doubling the initial distance the weight is displaced from equilibrium doubles the amplitude, but does not change the period or the phase of the oscillator. So in the general equation of an oscillator:

$x(t)=Acos(\omega t+\phi)$ , only the term changes, and it doubles.

Springs

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