All High School Physics Resources
Example Questions
Example Question #883 : High School Physics
A spring is attached to a mass, oscillating freely in simple harmonic motion. What change can be made to increase the period of the oscillation?
Decrease the mass at the end of the spring
Take the spring to a planet with a lower acceleration due to gravity
Increase the mass at the end of the spring
Use a spring with a greater spring constant
Lengthen the spring
Increase the mass at the end of the spring
The equation for the period of a spring in simple harmonic motion is:
In this formula, m is the mass and k is the spring constant. The only two things we can adjust that can change the period, then, are the mass and spring constant. The length of the spring and the acceleration due to gravity are irrelevant.
If we increase the mass we get a larger numerator, which in turns will give us a larger period. If we decrease the mass we get a smaller numerator, which would give us a smaller period. If we use a higher spring constant we get a larger denominator, which also gives us a smaller period.
Example Question #1 : Period Of Shm
A man swings a bucket back and forth at the end of a rope, creating a pendulum. What factor could be used to increase the period of the pendulum?
Move the pendulum to a lower altitude
Add water to the bucket
Use a stronger rope
Increase the length of the rope
Remove water from the bucket
Increase the length of the rope
The equation for the period of a pendulum is:
Notice that the material of the pendulum and the mass at the end do not enter into the equation at all. The only things that are capable of affecting the period are the length of the pendulum and the acceleration due to gravity.
While changing the altitude of the pendulum will change the period, it will do it only slightly unless you take it miles above the earth's crust. Furthermore, decreasing the altitude of the pendulum will increase the force due to gravity, which will result in a decreased period. The best answer is to increase the length of the rope. This will increase the period of the pendulum.
Example Question #884 : High School Physics
Your grandfather clock’s pendulum has a length of . If the clock loses half a minute per day, how should you adjust the length of the pendulum?
We should lengthen the pendulum by
We should lengthen the pendulum by
We should length the pendulum by
We should shorten the pendulum by
We should shorten the pendulum by
We should lengthen the pendulum by
We also can calculate the total number of seconds in a day.
There are seconds in one day.
Therefore we want our clock to swing a certain number of times with a period of to equal .
We know that our current clock has a certain number of swings with a period of to equal
So we have
We can calculate the current period of the pendulum using the equation
We can set up a ratio of each of these two periods to determine the missing length.
Notice that 2, pi and g are all in both the numerator and denominator and therefore fall out of the problem.
We can now solve for our missing piece.
Square both sides to get rid of the square root.
We should lengthen the pendulum by
Example Question #1 : Period Of Shm
A mass is placed at the end of a spring. It has a starting velocity of and is allowed to oscillate freely. If the mass has a starting velocity of , what would the period be?
The period would be 4 times as long.
The period would be twice as long.
The period would remain the same.
The period would be 14 as long.
The period would be half as long.
The period would remain the same.
When it comes to the period of a spring, the velocity of the object has no effect.
The equation is
Example Question #3 : Period Of Shm
A spring has a mass attached to one, which oscillates with a period of . What is the frequency?
The mass has no bearing on the relationship between frequency and period. This relationship is given by the equation:
Given the period, the frequency will be equal to its reciprocal.
Example Question #6 : Period Of Shm
A pendulum has a period of on Earth. What is its period on Mars, where the acceleration due to gravity is about that on Earth?
To begin we want to determine the original length of the pendulum.
The equation to determine the period of pendulum is
We can rearrange this equation to solve for length ().
We can now solve for the length knowing that the period of the pendulum is 2 seconds.
Now we can use the same equation and instead substitute in the value for the acceleration due to gravity on Mars.
Example Question #7 : Period Of Shm
How long must a simple pendulum be if it is to make exactly one swing per second? (That is one complete oscillation would take 2 seconds.)
The equation to determine the period of pendulum is
We can rearrange this equation to solve for length ().
We can now solve for the length knowing that the period of the pendulum is 2 seconds.
Example Question #1 : Period Of Shm
If a particle undergoes Simple Harmonic motion with an amplitude of . What is the total distance it travels in one period?
The period is how long it takes to complete one oscillation. In this case, a complete oscillation would take the object away from the center (the amplitude), travel back to the center, travel below the equilibrium point away, and then travel back to the center. Therefore the total distance traveled is times the amplitude as it must travel away, back, below, and back again. Therefore the total distance traveled is .
Example Question #391 : Motion And Mechanics
What is the period of a simple pendulum long when it is in a freely falling elevator?
Infinite
Infinite
The period of a simple pendulum on the free-falling is infinite because at a time of free-falling elevator the gravitational constant will equal zero which leads to an increase in the value in the numerator to infinite. Thus, the period of time in a free-falling elevator is infinite.
Example Question #1 : Energy Of Shm
A vertical spring with a spring constant of is stationary. An mass is attached to the end of the spring. What is the maximum displacement that the spring will stretch?
The best way to solve this problem is by using energy. Notice that the spring on its own is stationary. That means its initial total energy at that moment is zero. When the mass is attached, the spring stretches out, giving it spring potential energy (PEspring).
Where does that energy come from? The only place it can come from is the addition of the mass. Since the system is vertical, this mass will have gravitational potential energy.
Use the law of conservation of energy to set these two energies equal to each other:
We are trying to solve for displacement, and now we have an equation in terms of our variable.
Start by diving both sides by Δy to get rid of the Δy2 on the right side of the equation.
We are given values for the spring constant, the mass, and gravity. Using these values will allow use to solve for the displacement.
Note that the displacement will be negative because the spring is stretched in the downward direction due to gravity.