Harmonic Motion

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AP Physics 1 › Harmonic Motion

Questions 1 - 10

A 30 kg weight attached to a spring is at equilibrium lying horizontally on a table. The spring is lifted up and is stretched by 80 cm before the weight is lifted off the table. What is this spring's spring constant ()?

$368\frac{kg}{s^2}$

$3.68\frac{kg}{s^2}$

$0.375\frac{kg}{s^2}$

$37.5\frac{kg}{s^2}$

$460\frac{kg}{s^2}$

Explanation

First, we should convert 80 centimeters to 0.8 meters.

We know the force applied to the weight by gravity is

$294N=(30kg)(9.81\frac{m}{s^2})$

The force applied by the spring in the opposite direction must be equal to this:

$k=368\frac{N}{m}=368\frac{kg}{s^2}$

$368\frac{kg}{s^2}$

$3.68\frac{kg}{s^2}$

$0.375\frac{kg}{s^2}$

$37.5\frac{kg}{s^2}$

$460\frac{kg}{s^2}$

Explanation

First, we should convert 80 centimeters to 0.8 meters.

We know the force applied to the weight by gravity is

$294N=(30kg)(9.81\frac{m}{s^2})$

The force applied by the spring in the opposite direction must be equal to this:

$k=368\frac{N}{m}=368\frac{kg}{s^2}$

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

The student then replaces the weight with a weight whose mass, is twice as large as that of the original weight without changing the length of the string. The student again releases the weight from rest from the same displacement from equilibrium. What would the new equation of motion be for the pendulum?

$x(t)=0.13m;cos(6.3\frac{radians}{s}t+0.14:radians)$

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

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

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

$x(t)=0.13m;cos(3.15\frac{radians}{s}t+0.14:radians)$

Explanation

The period and frequency of a pendulum depend only on its length and the gravity force constant, . Changing the mass of the pendulum does not affect the frequency, and since the student released the new pendulum from the same displacement as the old, the amplitude and phase remain the same, and the equation of motion is the same for both pendula.

$x(t)=0.13m;cos(6.3\frac{radians}{s}t+0.14:radians)$

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

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

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

$x(t)=0.13m;cos(3.15\frac{radians}{s}t+0.14:radians)$

Explanation

You push on a door with a force of 1.3 N at an angle of 45 degrees to the surface and at a distance of 0.5 m from the hinges. What is the torque produced?

$0.46N\cdot m$

$1.2N\cdot m$

$0.31N\cdot m$

$0.84N\cdot m$

$0.64N\cdot m$

Explanation

To calculate torque, this equation is needed:

$\tau=rFsin\Theta$

Next, identify the given information:

$\Theta=45^{o}$

Plug these numbers into the equation to determine the torque:

$\tau=rFsin\Theta = (0.5m)(1.5N)sin(45^{o})=0.46N\cdot m$

You push on a door with a force of 1.3 N at an angle of 45 degrees to the surface and at a distance of 0.5 m from the hinges. What is the torque produced?

$0.46N\cdot m$

$1.2N\cdot m$

$0.31N\cdot m$

$0.84N\cdot m$

$0.64N\cdot m$

Explanation

To calculate torque, this equation is needed:

$\tau=rFsin\Theta$

Next, identify the given information:

$\Theta=45^{o}$

Plug these numbers into the equation to determine the torque:

$\tau=rFsin\Theta = (0.5m)(1.5N)sin(45^{o})=0.46N\cdot m$

Which simple pendulum will have a longer period?

$A:m = 10\textup{ kg}, L = 2\textup{ m}$

$B: m = 4\textup{ kg}, L = 3\textup{ m}$

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

B, because it has a longer length

A, because it has a shorter length

B, because it has a smaller mass

A, because it has a larger mass

They will have the same period

Explanation

The expression for the period of a pendulum is:

$T = 2\pi \sqrt{\frac{L}{g}}$

Therefore, the period of a pendulum is proportional to the square root of the length of the pendulum (assuming they are both on earth, or the same planetary body). Thus, the pendulum with the longer length will have the longer period.

A horizontal spring with a constant $k = 250\ \frac{\textup{N}}{\textup{m}}$ is on a frictionless surface. If the mass is doubled, by what factor is the frequency of the spring changed? Assume simple harmonic motion.