1
Quiz 1 solutions are in order of the questions for version A.
(1). Correct answer A
The lowest frequency for a string clamped at both ends is f = c/2L
(2). Correct answer is B
345
This problem involves two Doppler shifts. First, the wall hear a sound fwall = f0 345−17
with f0 = 345/λ and then the reflected sound is heard with frequency f = fwall 345+17
.
345
Thus
345
∆f ≡ f − f0 =
.15
362
− 1 ' 240Hz
328
(3). Correct answer is A
When the string creates a vibration in the air, they must have the same frequency
(4). Correct answer is D
Waves do not transport matter. All the other statements are true
(5). Correct answer is A
The frequency of organ pipe modes (open-open condition) is
fn = n
v
2L
Hence the difference between any two successive modes is just 345/2L = 30Hz; solving
for L gives answer A
(6). Correct answer is A
The path difference δ must equal half a wavelength. We get the correct answer from
δ=
p
p
8.32 + (2.26 + 2.11)2 − 8.32 + (2.26 − 2.11)2
(7). Correct answer is A
The period of time between intensity maxima is just 1/(beat frequency). Here the
beat frequency is just the difference of the two frequencies, i.e. 38/min = 38/60 Hz
(8). Correct answer is D
The tension is just F= µv 2 , where µ is the mass per unit length. This can determined
from the density by multiplying with the cross sectional area; hence
F = 2700 ∗ π ∗ (2.3 × 10−3 )2 ∗ (30)2 ' 40N
2
(9). Correct answer is D
Energy transport is always proportional to the square of the amplitude
(10). Correct answer is D
The fact that sound intensity goes down from he source is a direct consequence of
conservation of energy