Waves Practice Problems for AP Physics B & C
Review the following concepts if necessary:
- Transverse and Longitudinal Waves for AP Physics B & C
- Waves Interference for AP Physics B & C
- Standing Waves for AP Physics B & C
- Doppler Effect and Electromagnetic Waves for AP Physics B & C
- Single and Double Slits for AP Physics B & C
- Index of Refraction for AP Physics B & C
- Waves: Of Special Interest to Physics C Students
- A talk show host inhales helium; as a result, the pitch of his voice rises. What happens to the standing waves in his vocal cords to cause this effect?
- The wavelength of these waves increases.
- The wavelength of these waves decreases.
- The speed of these waves increases.
- The speed of these waves decreases.
- The frequency of these waves decreases.
- In a pipe closed at one end and filled with air, a 384-Hz tuning fork resonates when the pipe is 22-cm long; this tuning fork does not resonate for any smaller pipes. For which of these closed pipe lengths will this tuning fork also resonate?
- 11 cm
- 44 cm
- 66 cm
- 88 cm
- 384 cm
- Monochromatic light passed through a double slit produces an interference pattern on a screen a distance 2.0 m away. The third-order maximum is located 1.5 cm away from the central maximum. Which of the following adjustments would cause the third-order maximum instead to be located 3.0 cm from the central maximum?
- doubling the distance between slits
- tripling the wavelength of the light
- using a screen 1.0 m away from the slits
- using a screen 3.0 m away from the slits
- halving the distance between slits
- The two wave pulses shown above are moving toward each other along a string. When the two pulses interfere, what is the maximum amplitude of the resultant pulse?
- Laser light is passed through a diffraction grating with 7000 lines per centimeter. Light is projected onto a screen far away. An observer by the diffraction grating observes the first order maximum 25° away from the central maximum.
- What is the wavelength of the laser?
- If the first order maximum is 40 cm away from the central maximum on the screen, how far away is the screen from the diffraction grating?
- How far, measured along the screen, from the central maximum will the second-order maximum be?
- C—The frequency of these waves must go up, because the pitch of a sound is determined by its frequency. The wavelength of the waves in the host's voice box doesn't change, though, because the wavelength is dependent on the physical structure of the host's body. Thus, by v = λf, the speed of the waves in his vocal cords must go up. You can even look it up—the speed of sound in helium is faster than the speed of sound in normal air.
- C—A wave in a pipe closed at one end has a node at one end and an antinode at the other. 22 cm is the length of the pipe for the fundamental oscillation, which looks like this:
- E—Use the equation
- D—When wave pulses interfere, their amplitudes add algebraically. The question asks for the maximum amplitude, so the widths of the pulses are irrelevant. When both pulses are right on top of one another, each pulse will have amplitude A; these amplitudes will add to a resultant of 2A.
- (a) Use d sin θ = mλ. Here d is not 7000! d represents the distance between slits. Because there are 7000 lines per centimeter, there's 1/7000 centimeter per line; thus, the distance between lines is 1.4 × 10–4 cm, or 1.4 × 10–6 m. θ is 25° for the first-order maximum, where m = 1. Plugging in, you get a wavelength of just about 6 × 10–7 m, also known as 600 nm.
You can see that 1/2 of a "hump" is contained in the pipe, so the total wavelength (two full "humps") must be 88 cm. The next harmonic oscillation occurs when there is again a node at one end of the pipe and an antinode at the other, like this:
This pipe contains 11/2 "humps," so its length equals three-quarters of the total wavelength. The pipe length is thus 66 cm.
Here m = 3 because we are dealing with the third-order maximum. We want to double the distance to this third-order maximum, which means we want to double x in the equation. To do this, halve the denominator; d in the denominator represents the distance between slits.
(b) This is a geometry problem. tan 25° = (40 cm)/L; solve for L to get 86 cm, or about 3 feet.
(c) Use d sin θ = mλ; solve for θ using m = 2, and convert everything to meters. We get sin θ = 2(6.0 × 10–7 m)/(1.4 × 10–6 m). The angle will be 59°. Now, use the same geometry from part (b) to find the distance along the screen: tan 59° = x/(0.86 m), so x = 143 cm. (Your answer will be counted correct if you rounded differently and just came close to 143 cm.)