Student _________________________________
1. (a)
AP Physics 1
Speed of Waves – Answer Key
Date ___________
As one domino topples against the next one in line, it is moving partly perpendicular and
partly parallel to the direction along which the disturbance propagates.
2. 4A
The amplitude specifies the maximum excursion of the spot from the spot’s undisturbed
position, and the spot moves through this distance four times during each cycle. For
instance, in one cycle starting from its undisturbed position, the spot moves upward a
distance A, downward a distance A (returning to its undisturbed position), downward again a
distance A, and finally upward again a distance A (returning to its undisturbed position).
3. The amplitude specifies the maximum displacement of a particle from its undisturbed
position and has nothing to do with the wave speed.
4. (c)
The speed of the wave is greater when the tension in the rope is greater, which it is in the upper
portion of the rope. The rope has a mass m and, hence, a weight. The upper portion of the rope
has a greater tension than does the lower portion, because the upper portion supports the weight
of a greater length of rope hanging below.
5. (b)
Sound travels faster in liquids than in gases. The greater speed in water ensures that the
echo will return more quickly in water than in air.
6. The frequency of the sound is determined by the vibrating diaphragm of the horn. The
sound wave travels through the air and contacts the surface of the water, where it causes the
water molecules to vibrate at the same frequency as the molecules in the air. However, the
speed of sound in air is smaller than in air. Therefore, the wavelength is proportional to the
speed, when the frequency is constant. As a result, the wavelength is smaller in the air than in the
water.
7. a. After the initial crest passes, 5 additional crests pass in a time of 50.0 s. The period T of
the wave is: T= 50.0 s /5 = 10.0 s
b. Since the frequency f and period T are related by f = 1/T, we have: f=1/10.0s = 0.100 Hz
c. The horizontal distance between two successive crests is given as 32 m. This is also the
wavelength λ of the wave, so λ = 32 m
d. The speed v of the wave is v = fλ = (0.100 Hz)(32 m) = 3.2 m/s
e. There is no information given, either directly or indirectly, about the amplitude of the
wave. Therefore, it is not possible to determine the amplitude.
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8.
𝐹𝐹𝑇𝑇
The speed v of a transverse wave on a wire is given by 𝑣𝑣 = � 𝑚𝑚 , where FT is the tension
𝐿𝐿
and m/L is the mass per unit length (or linear density) of the wire. We are given that FT and
m are the same for the two wires, and that one is twice as long as the other. This
information, along with knowledge of the wave speed on the shorter wire, will allow us to
determine the speed of the wave on the longer wire.
𝐹𝐹𝑇𝑇
𝐹𝐹
𝑣𝑣𝑠𝑠ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 = � 𝑚𝑚𝑇𝑇
The speeds on the longer and shorter wires are: 𝑣𝑣𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 = � 𝑚𝑚
2𝐿𝐿
Dividing these two expressions side by side gives:
𝑣𝑣𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
=
𝑣𝑣𝑠𝑠ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜
𝐹𝐹𝑇𝑇
� 𝑚𝑚
2𝐿𝐿
𝐹𝐹𝑇𝑇
� 𝑚𝑚
𝐿𝐿
𝐿𝐿
= √2
𝑣𝑣𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 = √2𝑣𝑣𝑠𝑠ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 = 340 𝑚𝑚/𝑠𝑠
9. The length L of the string is one of the factors that affects the speed of a wave traveling on it,
in so far as the speed v depends on the mass per unit length m/L according to the formula
𝐹𝐹
𝑣𝑣 = � 𝑚𝑚𝑇𝑇 . The other factor affecting the speed is the tension FT. The speed is not directly given
𝐿𝐿
here. However, the frequency f and the wavelength λ are given, and the speed is related to them
according to v = f λ. Combining the two equations and solving for L we get:
𝐹𝐹
𝑓𝑓 𝜆𝜆 = � 𝑚𝑚𝑇𝑇
𝐿𝐿 =
𝐿𝐿
𝑓𝑓 2 𝜆𝜆2 𝑚𝑚
𝐹𝐹𝑇𝑇
= 0.68 𝑚𝑚
𝐹𝐹𝑇𝑇
10. The speed v of the waves on the strand of silk is given by 𝑣𝑣 = � 𝑚𝑚 . The tension FT is
𝐿𝐿
directly proportional to the spider’s mass M, because the tension force exerted on the spider by
the silk is equal in magnitude to the spider’s weight Mg, where g is the magnitude of the
acceleration due to gravity: 𝐹𝐹𝑇𝑇 = 𝑀𝑀𝑀𝑀. Neither the mass m nor the length L of the silk strand are
given, but we know the density ρof the silk, which is the ratio of the mass m of the strand to its
volume V, according to ρ = . The strand is a cylinder, so its volume V is the product of its
𝑉𝑉
length L and its cross-sectional area A: 𝑉𝑉 = 𝐴𝐴𝐴𝐴 = 𝜋𝜋𝑟𝑟 2 𝐿𝐿. Substituting 𝐹𝐹𝑇𝑇 = 𝑀𝑀𝑀𝑀 we get:
𝑚𝑚
𝑣𝑣 = �
Substituting:
𝑀𝑀𝑀𝑀
𝑚𝑚
𝐿𝐿
𝑚𝑚
𝐿𝐿
= 𝜋𝜋ρ𝑟𝑟2
and solving for M:
we get:
𝑀𝑀 =
𝑀𝑀 =
𝑣𝑣 2 (𝜋𝜋ρ𝑟𝑟2 )
𝑔𝑔
𝑚𝑚
𝐿𝐿
𝑣𝑣 2 ( )
𝑔𝑔
= 5.2×10−4 𝑘𝑘𝑘𝑘
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