AP Physics 1 : Period and Frequency

Study concepts, example questions & explanations for AP Physics 1

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Example Questions

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Example Question #1072 : Newtonian Mechanics

A ferris wheel has a diameter of 30 meters. The carriages on the outside of the wheel are traveling at an instantaneous velocity of . What is the period of rotation of the wheel?

Possible Answers:

Correct answer:

Explanation:

We can calculate the circumference of the wheel using the given radius:

We can use this and the given velocity to find the period:

Example Question #1073 : Newtonian Mechanics

A mass of  is traveling in a circle of . If it is under a centripetal force of , what is the mass's period?

Possible Answers:

Correct answer:

Explanation:

Knowing the centripetal force on the mass and the radius of the circle, we can calculate its velocity:

Rearranging for velocity:

We can use this to find the period of the mass:

Rearranging for period, we get:

Example Question #1074 : Newtonian Mechanics

A grandfather clock is getting extremely rusty on the inside and falls behind by 1500 seconds every day. What is the period of the minute hand of the clock?

Possible Answers:

Correct answer:

Explanation:

A normal clock registers 60 seconds for every rotation of the minute hand. We need to determine the new period per rotation.

A normal clock registers the following amount of seconds every day:

Losing 1500 seconds every day, we now have:

The hand in question completes one rotation every minute, so we'll dive this new time by minutes per day:

Now dividing our two values to get seconds per rotation:

Example Question #1075 : Newtonian Mechanics

Suppose you have a string of length  with a ball of mass  attached to the end. You are going to spin the ball in a vertical circle. What is the minimum frequency of the ball that will keep the string in tension at all times?

Possible Answers:

Correct answer:

Explanation:

To start, we need to determine what exactly we are solving for. What does it mean for the string to be in tension at all points? This means that at some point in the circle, tension will equal zero; thus the force of gravity and the centripetal force will equal each other.

Expand these force expressions and simplify:

The expression for centripetal acceleration is:

We know  (the length of the string) and , but we need to develop an expression for velocity:

Substituting this back into the equation for centripetal acceleration, we get:

Rearrange for frequency:

We know all of these values, allowing us to solve:

Example Question #1 : Period And Frequency

An astronaut in space has a ball of mass  attached to the end of a string of length . The ball is spun in a horizontal circle. If the string breaks under a force of , what is the minimum period at which the ball can be spun?

Possible Answers:

Correct answer:

Explanation:

Since the man is in space, the only force we have to worry about in this problem is the centripetal force, which results from the tension in the spring. Therefore, we are being asked what period gives us a centripetal force of .

We need an expression for velocity:

Substitute this back into the original expression:

Rearranging for period we get:

We know all of these values, allowing us to solve:

Example Question #2 : Period And Frequency

What is the ordinary frequency of the second hand on a clock?

Possible Answers:

Correct answer:

Explanation:

The ordinary frequency is the number of cycles per second. Since a second hand makes one revolution, or cycle, every 60s, the correct answer is . You can also think of the ordinary frequency as the angular velocity divided by .

Example Question #121 : Circular And Rotational Motion

A solid cylinder of mass  and radius  is at rest at the top of a slope with an angle of . The sphere is then released. How far down the slope has the sphere traveled when it has a period of . Neglect air resistance and any frictional forces.

Possible Answers:

Correct answer:

Explanation:

We can begin with the conservation of energy to solve this problem:

The problem statement tells us that the cylinder is initially at rest, so we can eliminate initial kinetic energy. If we assume that the height of the cylinder when it reaches a period of 0.2s has a height of 0, we can eliminate final potential energy. Therefore, we get:

Expanding these terms and making sure we have both a linear and rotational component to kinetic energy, we get equation (1):

Before we move on, we know that we are going to have to calculate something that we can use to determine the period of the cylinder. We know that the period is how long it takes the cylinder to complete one full rotation. Thinking practically, we can use the circumference of the cylinder and linear velocity to determine period:

Using variables, we get equation:

Rearranging for final velocity, we get equation (2):

Now we know that the period is dependent on final linear velocity. We will come back to this equation. Now we can go back to equation (1) and begin substituting in expressions for unknown variables moving from left to right. The first variable we don't know is initial height. However, we can use the distance the cylinder traveled and the angle of the slope:

Rearranging for initial height, we get equation (3):

Moving on, the next unknown term is final velocity. We can substitute equation (2) that we already derived:

Moving on, the next unknown term is the moment of inertia. Using the expression for a cylinder to get equation (4):

Moving on, the final unknown term is final rotational velocity. We can use the relationship between this and linear velocity:

Now substituting equation (2), we get equation (5):

We can now substitute equations 2, 3, 4, and 5 into equation (1):

Eliminating mass from both sides of the equation and expanding each term:

Combining the terms on the right:

Rearranging for length:

Check your units and make sure you end up with seconds before moving on!

We know values for each variable, so time to plug and chug:

Example Question #122 : Circular And Rotational Motion

Two cars are racing side by side on a perfectly circular race track. The inner car is from the center of the track. The outer car is from center of the track. Both cars are traveling at .

How long does it take the inner car to complete a lap?

Possible Answers:

Correct answer:

Explanation:

Finding distance of lap of inner car:

Convert  to :

Use the distance formula:

Example Question #123 : Circular And Rotational Motion

Two cars are racing side by side on a perfectly circular race track. The inner car is from the center of the track. The outer car is from center of the track. Both cars are traveling at 

How much longer does it take the outer car to complete one lap?

Possible Answers:

Correct answer:

Explanation:

Inner car time:

Finding distance of lap of inner car:

Convert  to :

Use the distance formula:

Finding distance of lap of outer car:

Convert  to :

Use the distance formula:

Example Question #3 : Period And Frequency

A car with wheels of mass and wheels of radius is traveling at . Treating the wheels as disks of uniform mass density, calculate the angular frequency of one wheel.

Possible Answers:

None of these

Correct answer:

Explanation:

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