Periodic Motion and Waves

Question 1

Objects in periodic motion are in sync when…

Answer 1

Objects in periodic motion are in sync when they reach their respective maximum amplitude at the same times

Question 2

True or false: The force of gravity is greatest when the pendulum is at its equilibrium position.

Answer 2

This statement is false. The force of gravity is constant throughout the pendulum’s motion. The force of gravity is dependent on the mass of the pendulum and the acceleration due to gravity, both of which remain unchanged.

Question 3

How do you calculate the period of a spring?

Answer 3

Equation:

Question 4

How do you calculate the period of a pendulum?

Answer 4

Equation:

Question 5

The number of times the pendulum swings back and forth in one second

Answer 5

frequency refers to the number of times the pendulum swings back and forth in a second.

Question 6

The amount of time needed for the pendulum to swing back and forth once

Answer 6

Period

Question 7

The maximum displacement of the pendulum from its equilibrium position

Answer 7

Amplitude

Question 8

Oscillating electromagnetic fields

Answer 8

Electromagnetic wave

Question 9

The movement of particles is perpendicular to the movement of the wave

Answer 9

Transverse wave

Question 10

Movement of physical objects back and forth

Answer 10

Mechanical wave

Question 11

the particles are moving in parallel with the wave

Answer 11

Longitudinal wave

Question 12

If you combined these waves, where would you see the greatest amount of constructive interference?

Answer 12

The point with the highest amplitude will occur when there is the greatest constructive interference. At point C, both graphs are close to being at their peak, so the combined amplitude will be the largest at this point.

Question 13

The wavelength of the third harmonic of an open tube is 1.5 meters. What is the length of the tube?

Answer 13

The formula for the wavelength of an open tube is λ=2L/n. Plugging in the values from the question results in:

1. 5=(2L)/3
2. 5=2L
3. 25=L.

The units are meters since the wavelength is given in meters.

Question 14

Alex plucks a 75 cm fixed string and hears the second harmonic. What frequency does Alex hear? Use 340 m/s as the speed of sound.

Answer 14

The wavelength of the sound is:

λ=(2L)/n

λ=(2(.75))/2

λ=.75 m

Note that we have to convert from centimeters to meters since the speed of sound is given in meters per second. Using v=fλ,

V=fλ

340 m/s = f(.75 m)

453Hz=f

Question 15

The fourth harmonic for strings has _______ while the fourth harmonic for open pipes has _______.

Answer 15

For fixed strings, the number of antinodes is the same value as the number of harmonics. For open pipes, the number of harmonics is reflective of the number of nodes. Both share the same equation for the wavelength (λ=(2L)/n).

Question 16

True or false: in a closed tube, if the wavelength of the second harmonic is 20 cm, then the length of the tube is 10 cm.

Answer 16

This statement is false. For closed tubes, there can only be odd number harmonics.

Question 17

If Matthew blew air into a 1 meter closed tube, what will be the frequency of the third harmonic?

Answer 17

The formula for the wavelength of a closed tube is λ=(4L)/n. Plugging in the values from the question:

λ=(4(1))/3

λ=4/3

From here, we can find the frequency via

V=fλ

340=f(4/3)

255 hz=f

Question 18

Standing wave

Answer 18

Standing waves are created from patterns of constructive or destructive interference between multiple other waves.

Question 19

True or False: Resonance frequencies are an example of constructive interference

Answer 19

Resonance frequencies are an example of constructive interference because they increase the amplitude.

Question 20

If the frequency of a block on a string is 25 Hz, its period is:

Answer 20

The period is the reciprocal of the frequency. If the frequency is 25Hz, then the period is 1/(25Hz), or 0.04 seconds.

Question 21

As a sound wave enters a new medium, the velocity decreases. In response, the period:

Answer 21

The frequency remains the same between mediums. The change in velocity is reflected in a change in wavelength. Since the frequency remains constant, the period also remains constant.

Question 22

A plucked 1 meter string has a frequency of 680 Hz. How many antinodes does the corresponding harmonic have? (Assume that the speed of sound on the string is 340 m/s.)

Answer 22

First, we need to find the wavelength by using v = f*λ

340 = 680*λ

1/2= λ

Using the wavelength, we can find the number of harmonics by using λ=(2L)/n.

λ=(2L)/n

1/2=(2*1)/n

4=n

For fixed strings, the number of antinodes is the same as the number of harmonics.

Question 23

There is a box attached to a spring. Daniel switches the spring for a stiffer spring and compares the period of the box’s oscillation on this spring to its period on the old one. He should expect:

Answer 23

The stiffer spring has a greater spring constant. Recall that the formula for a spring’s period is 2π*sqrt(m/k). A greater spring constant, k, would result in a smaller value inside the square root. This would decrease the period.

Question 24

True or False: Sound waves are transverse waves.

Answer 24

This statement is false. Sound waves are longitudinal waves, meaning the particles of the medium move parallel to the direction that the wave moves.

Question 25

How do you calculate the velocity of a wave?

Answer 25

V=wavelength*frequency

Question 26

Describe a period in wave motion

Answer 26

The wave period is the time it takes to complete one cycle. The standard unit of a wave period is in seconds, and it is inversely proportional to the frequency of a wave, which is the number of cycles of waves that occur in one second. In other words, the higher the frequency of a wave, the lower the wave period.

Question 27

How do you calculate the wavelength of a standing wave of a string?

Answer 27

lamba= 2L/n

Where L is the length of the string and n is the harmonic

Question 28

How do you calculate the wavelngth of an open pipe?

Answer 28

lamba= 2L/n

Where L is the length of the open pipe and n is the harmonic

Question 29

How do you calculate the wavelngth of a closed pipe?

Answer 29

Lambda=4L/n

Where lamba is wavelngth, L is length of the closed pipe, and n is the harmonic

Question 30

If ocean waves strike the shore every 3.0 s and the horizontal distance between adjacent crests and troughs is 1.0 m, what is the average speed of the waves?

Answer 30

0.67 m/s