Pendulums - AP Physics 1
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A ball of mass 2kg is attached to a string of length 4m, forming a pendulum. If the string is raised to have an angle of 30 degrees below the horizontal and released, what is the velocity of the ball as it passes through its lowest point?
A ball of mass 2kg is attached to a string of length 4m, forming a pendulum. If the string is raised to have an angle of 30 degrees below the horizontal and released, what is the velocity of the ball as it passes through its lowest point?
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This question deals with conservation of energy in the form of a pendulum. The equation for conservation of energy is:
According to the problem statement, there is no initial kinetic energy and no final potential energy. The equation becomes:
Subsituting in the expressions for potential and kinetic energy, we get:
We can eliminate mass to get:
Rearranging for final velocity, we get:
In order to solve for the velocity, we need to find the initial height of the ball.
The following diagram will help visualize the system:
From this, we can write:
Using the length of string and the angle it's held at, we can solve for 
:
Now that we have all of our information, we can solve for the final velocity:
This question deals with conservation of energy in the form of a pendulum. The equation for conservation of energy is:
According to the problem statement, there is no initial kinetic energy and no final potential energy. The equation becomes:
Subsituting in the expressions for potential and kinetic energy, we get:
We can eliminate mass to get:
Rearranging for final velocity, we get:
In order to solve for the velocity, we need to find the initial height of the ball.
The following diagram will help visualize the system:
From this, we can write:
Using the length of string and the angle it's held at, we can solve for
Now that we have all of our information, we can solve for the final velocity:
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: 
The student then replaces the string with a string whose length, 
is twice as large as that of the original string without changing the mass of the weight. 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?
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:
The student then replaces the string with a string whose length,
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Doubling the length of a pendulum increases the period, so it decreases the frequency of the pendulum. The frequency depends upon the square root of the length, so the frequency decreases by a factor of 
. Neither of the other parameters (amplitude, phase) change.
Doubling the length of a pendulum increases the period, so it decreases the frequency of the pendulum. The frequency depends upon the square root of the length, so the frequency decreases by a factor of
A pendulum has a period of 5 seconds. If the length of the string of the pendulum is quadrupled, what is the new period of the pendulum?
A pendulum has a period of 5 seconds. If the length of the string of the pendulum is quadrupled, what is the new period of the pendulum?
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We need to know how to calculate the period of a pendulum to solve this problem. The formula for period is:
In the problem, we are only changing the length of the string. Therefore, we can rewrite the equation for each scenario:
Dividng one expression by the other, we get a ratio:
We know that 
, so we can rewrite the expression as:
Rearranging for P2, we get:
We need to know how to calculate the period of a pendulum to solve this problem. The formula for period is:
In the problem, we are only changing the length of the string. Therefore, we can rewrite the equation for each scenario:
Dividng one expression by the other, we get a ratio:
We know that
Rearranging for P2, we get:
A student studying Newtonian mechanics in the 19th century was skeptical of some of Newton's concepts. The student has a pendulum that has a period of 3 seconds while sitting on his desk. He attaches the pendulum to a ballon and drops it off the roof of a university building, which is 20m tall. Another student realizes that the pendulum strikes the ground with a velocity of 
. What is the period of the pendulum as it is falling to the ground?
Neglect air resistance and assume 
A student studying Newtonian mechanics in the 19th century was skeptical of some of Newton's concepts. The student has a pendulum that has a period of 3 seconds while sitting on his desk. He attaches the pendulum to a ballon and drops it off the roof of a university building, which is 20m tall. Another student realizes that the pendulum strikes the ground with a velocity of
Neglect air resistance and assume
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We need to know the formula for the period of a pendulum to solve this problem:
We aren't given the length of the pendulum, but that's ok. We could solve for it, but it would be an unnecessary step since the length remains constant.
We can write this formula for the pendulum when it is on the student's table and when it is falling:
1 denotes on the table and 2 denotes falling. The only thing that is different between the two states is the period and the gravity (technically the acceleration of the whole system, but this is the form in which you are most likely to see the formula). We can divide the two expressions to get a ratio:
Canceling out the constants and rearranging, we get:
We know g1; it's simply 10. However, we need to calculate g2, which is the rate at which the pendulum and balloon are accelerating toward the ground. We are given enough information to use the following formula to determine this:
Removing initial velocity and rearranging for acceleration, we get:
Plugging in our values:
This is our g2. We now have all of the values to solve for T2:
We need to know the formula for the period of a pendulum to solve this problem:
We aren't given the length of the pendulum, but that's ok. We could solve for it, but it would be an unnecessary step since the length remains constant.
We can write this formula for the pendulum when it is on the student's table and when it is falling:
1 denotes on the table and 2 denotes falling. The only thing that is different between the two states is the period and the gravity (technically the acceleration of the whole system, but this is the form in which you are most likely to see the formula). We can divide the two expressions to get a ratio:
Canceling out the constants and rearranging, we get:
We know g1; it's simply 10. However, we need to calculate g2, which is the rate at which the pendulum and balloon are accelerating toward the ground. We are given enough information to use the following formula to determine this:
Removing initial velocity and rearranging for acceleration, we get:
Plugging in our values:
This is our g2. We now have all of the values to solve for T2:
Consider the following system:
If the length of the pendulum is 
and the maximum velocity of the block is 
, what is the minimum possible value of angle A?
Consider the following system:
If the length of the pendulum is
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We can use the equation for conservation of energy to solve this problem.
If the initial state is when the mass is at its highest position and the final state is when the mass is at its lowest position, then we can eliminate initial kinetic energy and final potential energy:
Substituting expressions in for each term, we get:
Canceling out mass and rearranging for height, we get:
Thinking about a pendulum practically, we can write the height of the mass at any given point as a function of the length and angle of the pendulum:
Think about how this formula is written. The second term gives us how far down the mass is from the top point. Therefore, we need to subtract this from the length of the pendulum to get how high above the lowest point (the height) the mass currently is.
Substituting this into the previous equation, we get:
Rearrange to solve for the angle:
We have values for each variable, allowing us to solve:
We can use the equation for conservation of energy to solve this problem.
If the initial state is when the mass is at its highest position and the final state is when the mass is at its lowest position, then we can eliminate initial kinetic energy and final potential energy:
Substituting expressions in for each term, we get:
Canceling out mass and rearranging for height, we get:
Thinking about a pendulum practically, we can write the height of the mass at any given point as a function of the length and angle of the pendulum:
Think about how this formula is written. The second term gives us how far down the mass is from the top point. Therefore, we need to subtract this from the length of the pendulum to get how high above the lowest point (the height) the mass currently is.
Substituting this into the previous equation, we get:
Rearrange to solve for the angle:
We have values for each variable, allowing us to solve:
A pendulum of mass 
has a period 
. If the mass is quadrupled to 
, what is the new period of the pendulum in terms of 
?
A pendulum of mass
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The mass of a pendulum has no effect on its period. The equation for the period of a pendulum is
, which does not include mass.
The mass of a pendulum has no effect on its period. The equation for the period of a pendulum is
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: 
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?
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:
The student then replaces the weight with a weight whose mass,
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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.
The period and frequency of a pendulum depend only on its length and the gravity force constant,
A pendulum of length 
will take how long to complete one period of its swing?
A pendulum of length
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The period of a pendulum is given by the following formula:
Substituting our values, we obtain:
Roughly 6.3 seconds is the time it takes for the pendulum to complete one period.
The period of a pendulum is given by the following formula:
Substituting our values, we obtain:
Roughly 6.3 seconds is the time it takes for the pendulum to complete one period.
If a simple pendulum is set to oscillate on Earth, it has a period of 
. Now suppose this same pendulum were moved to the Moon, where the gravitational field is 6 times less than that of Earth.
What is the period 
of this pendulum on the Moon in terms of 
?
If a simple pendulum is set to oscillate on Earth, it has a period of
What is the period
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The period of a simple pendulum is given by:
Where 
is the period of the pendulum, 
is the length of the pendulum, and 
is the gravitational constant of the planet we are on. Thus on Earth, the period 
is given by:
With 
being Earth's gravitational constant. The period on the Moon is given by:
With 
being the Moon's gravitational constant. Since the Moon's gravity is 6 times weaker than that of Earth's, we have:
Plug this value into the Moon pendulum equation:
Since 
,
Substituting this into the above expression gives us 
The period of a simple pendulum is given by:
Where
With
With
Plug this value into the Moon pendulum equation:
Since
Substituting this into the above expression gives us
A pendulum of length 
has a mass of 
attached to the bottom. Determine the frequency of the pendulum if it is released from a shallow angle.
A pendulum of length
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The frequency of a pendulum is given by:
Where 
is the length of the pendulum and 
is the gravity constant. Notice how the frequency is independent of mass.
Plugging in values:
The frequency of a pendulum is given by:
Where
Plugging in values:
How will increasing the mass at the end of a pendulum change the period of it's motion? Assume a shallow angle of release.
How will increasing the mass at the end of a pendulum change the period of it's motion? Assume a shallow angle of release.
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The frequency of a pendulum is given by:
Where is the length of the pendulum and is the gravity constant. The frequency is independent of mass. Thus, adding mass will have no effect.
The frequency of a pendulum is given by:
Where
Which simple pendulum will have a longer period?
Which simple pendulum will have a longer period?
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The expression for the period of a pendulum is:
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.
The expression for the period of a pendulum is:
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 simple pendulum of length 
with a block of mass 
attached has a maximum velocity of 
. What is the maximum height of the block?
A simple pendulum of length
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This may come as a surprise, but we don't need to know a single formula concerning pendulums or circular motion to solve this problem. We just have to be able to understand the motion of a pendulum and think about the situation practically. A pendulum reaches its maximum velocity when the block is at its lowest point (the pendulum is vertical and pointing straight down). We can then use the expression for conservation of energy to determine the maximum height of the block.
If we assume that the low point of the pendulum has a height of 0, we can eliminate initial potential energy. We can also eliminate final kinetic energy (at least for now. If we get a maximum height that is more than twice the length of the pendulum, we know that it completes full rotations) since the block will be at rest when it reaches its maximum height.
Substituting expressions in for each term:
Eliminating mass and rearranging for final height, we get:
Plugging in our values, we get:
This may come as a surprise, but we don't need to know a single formula concerning pendulums or circular motion to solve this problem. We just have to be able to understand the motion of a pendulum and think about the situation practically. A pendulum reaches its maximum velocity when the block is at its lowest point (the pendulum is vertical and pointing straight down). We can then use the expression for conservation of energy to determine the maximum height of the block.
If we assume that the low point of the pendulum has a height of 0, we can eliminate initial potential energy. We can also eliminate final kinetic energy (at least for now. If we get a maximum height that is more than twice the length of the pendulum, we know that it completes full rotations) since the block will be at rest when it reaches its maximum height.
Substituting expressions in for each term:
Eliminating mass and rearranging for final height, we get:
Plugging in our values, we get:
A simple pendulum with a length of 
has a block of mass 
attached to the end. If the pendulum is 
above its lowest point and rotating downward, what is the instantaneous acceleration of the block?
A simple pendulum with a length of
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Since we are told that the block is 1m above its lowest point (half way between its low point and horizontal), we can calculate the angle that the pendulum makes with the vertical:
Then we can use the following expression to determine the net force acting on the block in the direction of its motion:
Then we can use Newton's 2nd law to determine the instantaneous acceleration:
Canceling out mass and rearranging for acceleration:
Since we are told that the block is 1m above its lowest point (half way between its low point and horizontal), we can calculate the angle that the pendulum makes with the vertical:
Then we can use the following expression to determine the net force acting on the block in the direction of its motion:
Then we can use Newton's 2nd law to determine the instantaneous acceleration:
Canceling out mass and rearranging for acceleration:
A block of mass 
is attached to a rigid pole of length 
. If the block has a velocity of 
as it travels through the horizontal, what is the distance between the blocks lowest and highest points? Neglect the mass of the pole. Neglect air resistance and any frictional forces.
A block of mass
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Since we know the velocity of the block as it travels through the horizontal, we can directly calculate the distance above horizontal that the block reaches using the expression for conservation of energy:
If we assume that horizontal has a height of 0, we can eliminate initial potential energy. We can also eliminate final kinetic energy as the block should be at rest when it reaches its highest point.
Plugging in expressions for each variable, we get:
Eliminating mass and rearranging for final height, we get:
Plugging in our values, we get:
This is the distance the block reaches above the horizontal, so we need to add this to the distance between horizontal and the blocks lowest point, which is simply the length of the pole:
Since we know the velocity of the block as it travels through the horizontal, we can directly calculate the distance above horizontal that the block reaches using the expression for conservation of energy:
If we assume that horizontal has a height of 0, we can eliminate initial potential energy. We can also eliminate final kinetic energy as the block should be at rest when it reaches its highest point.
Plugging in expressions for each variable, we get:
Eliminating mass and rearranging for final height, we get:
Plugging in our values, we get:
This is the distance the block reaches above the horizontal, so we need to add this to the distance between horizontal and the blocks lowest point, which is simply the length of the pole:
A rigid rod of length 
has a block of mass 
attached to one end and is allowed to rotate as a simple pendulum from the other end. What is the lowest maximum velocity of the block that will allow the block to rotate in complete circles?
A rigid rod of length
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The lowest maximum velocity occurs in the scenario where the block is at rest when the pendulum is vertical and pointing upward. Therefore, even the slightest additional movement will result in the pendulum rotating in complete circles With this in mind, we can begin with the expression for conservation of energy:
If we say the initial condition is when the block is at the lowest point of rotation (when it is traveling at its maximum velocity), and assume that point to have a height of 0, we can eliminate initial potential energy. Furthermore, our final state will then be when the block is at the highest point of rotation and at rest. Thus we can eliminate final kinetic energy to get:
Plugging in expressions for each variable, we get:
Eliminating mass and rearranging for maximum velocity, we get:
Where the maximum height is two times the length of the pendulum:
The lowest maximum velocity occurs in the scenario where the block is at rest when the pendulum is vertical and pointing upward. Therefore, even the slightest additional movement will result in the pendulum rotating in complete circles With this in mind, we can begin with the expression for conservation of energy:
If we say the initial condition is when the block is at the lowest point of rotation (when it is traveling at its maximum velocity), and assume that point to have a height of 0, we can eliminate initial potential energy. Furthermore, our final state will then be when the block is at the highest point of rotation and at rest. Thus we can eliminate final kinetic energy to get:
Plugging in expressions for each variable, we get:
Eliminating mass and rearranging for maximum velocity, we get:
Where the maximum height is two times the length of the pendulum:
A simple pendulum of length 
has a block of mass 
attached to the end of it. The pendulum is originally at an angle of 
to the vertical and at rest. If the pendulum is released and allowed to rotate freely at time 
, what is the angle of the pendulum at time 
?
A simple pendulum of length
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Since the maximum angle achieved by the pendulum is very small, we can use the follow expression to determine the angle of the pendulum at any time 
:
Note how we are using the cosine function since the pendulum began at its highest point. We already have all of these values, so we can simply plug and chug:
Since the maximum angle achieved by the pendulum is very small, we can use the follow expression to determine the angle of the pendulum at any time
Note how we are using the cosine function since the pendulum began at its highest point. We already have all of these values, so we can simply plug and chug:
A simple pendulum of length 
has a block attached to one end which has a maximum velocity of 
. What is the minimum velocity of the block?
A simple pendulum of length
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The block experiences its maximum velocity when it is at its lowest point of rotation and its minimum velocity at its highest point of rotation. Therefore, we can use the expression for conservation of energy to solve this problem:
If we assume that the initial condition is when the block is at the low point of rotation and assume that that point has a height of 0, then we can eliminate initial potential energy:
Now substituting in expressions for each of these:
Eliminating mass and multiplying both sides of the expression by 2, we get:
Then rearranging for final velocity:
Where the height is simply twice the length of the pendulum:
Plugging in our values:
The block experiences its maximum velocity when it is at its lowest point of rotation and its minimum velocity at its highest point of rotation. Therefore, we can use the expression for conservation of energy to solve this problem:
If we assume that the initial condition is when the block is at the low point of rotation and assume that that point has a height of 0, then we can eliminate initial potential energy:
Now substituting in expressions for each of these:
Eliminating mass and multiplying both sides of the expression by 2, we get:
Then rearranging for final velocity:
Where the height is simply twice the length of the pendulum:
Plugging in our values:
A simple pendulum has a length 
has a block of mass 
attached to one end. If the pendulum is released from rest, what is the maximum centripetal acceleration felt by the block?
A simple pendulum has a length
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Since we are given the length of the pendulum and told that it begins at rest and in the horizontal position, we can calculate the maximum velocity of the block as it travels through its lowest point using the expression for conservation of energy:
We can eliminate initial kinetic energy since the pendulum begins at rest. We can also eliminate final potential energy if we assume that the height at the lowest point of rotation is equal to 0.
Substituting in expressions for each of these:
Where the initial height is just the length of the pendulum:
Rearranging for final velocity, we get:
We can then calculate centripetal acceleration from this:
Where the radius is the length of the pendulum:
Since we are given the length of the pendulum and told that it begins at rest and in the horizontal position, we can calculate the maximum velocity of the block as it travels through its lowest point using the expression for conservation of energy:
We can eliminate initial kinetic energy since the pendulum begins at rest. We can also eliminate final potential energy if we assume that the height at the lowest point of rotation is equal to 0.
Substituting in expressions for each of these:
Where the initial height is just the length of the pendulum:
Rearranging for final velocity, we get:
We can then calculate centripetal acceleration from this:
Where the radius is the length of the pendulum:
A simple pendulum with a length 
has a block attached to one end that has maximum velocity of 
. At what angle to the vertical does the block have a velocity of 
?
A simple pendulum with a length
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We can begin with the expression for conservation of energy to solve this problem:
The block will achieve its maximum velocity at the lowest point of rotation. If we say that this point has a height of 0, we can eliminate initial potential:
Plugging in expressions for each of these:
Multiplying both sides of the expression by 
, we get:
Now let's say that the final condition is when the block has a velocity of 
. Rearranging for the final height, we get:
Plugging in our values:
This is the height that the block is above our reference point when it reaches the desired speed. From here, we can develop an expression for the height of the block as a function of the angle the block makes with the vertical. First, let's begin with a function that tells us how far below the block is from the horizontal:
If this does not make sense, draw it out. d is the distance below the horizontal, L is the length of the pendulum, and theta is the angle between the pendulum and the horizontal.
Moving on, we can take this expression to develop one that tells us the height of the block:
Rearranging for angle:
Now plugging in our values:
We can begin with the expression for conservation of energy to solve this problem:
The block will achieve its maximum velocity at the lowest point of rotation. If we say that this point has a height of 0, we can eliminate initial potential:
Plugging in expressions for each of these:
Multiplying both sides of the expression by
Now let's say that the final condition is when the block has a velocity of
Plugging in our values:
This is the height that the block is above our reference point when it reaches the desired speed. From here, we can develop an expression for the height of the block as a function of the angle the block makes with the vertical. First, let's begin with a function that tells us how far below the block is from the horizontal:
If this does not make sense, draw it out. d is the distance below the horizontal, L is the length of the pendulum, and theta is the angle between the pendulum and the horizontal.
Moving on, we can take this expression to develop one that tells us the height of the block:
Rearranging for angle:
Now plugging in our values: