A heavy stone of mass m is hung from the ceiling by a thin 8.25-g wire that is 65.0 cm long.When you gently pluck the upper end of the wire,a pulse travels down the wire and returns 7.84 ms later,having reflected off the lower end.The speed of sound in the room is 344 m/s,and the stone is heavy enough to prevent the lower end of the wire from moving.If the wire is vibrating in its second overtone,what is the frequency of the sound it will produce?
A) 128 Hz
B) 191 Hz
C) 255 Hz
D) 383 Hz
E) 765 Hz

Question

Quizplus · Accepted Answer

The pulse traveling down the wire and reflecting back will have traveled a round trip distance of 2L = 2(0.65 m) = 1.30 m. The time taken t for the round trip is given as 7.84 ms, which is equivalent to 7.84 × 10^-3 s. Therefore, the speed v of the pulse can be calculated as:

v = 2L/t = 2(0.65 m)/(7.84 × 10^-3 s) = 166.58 m/s

The frequency f of the wire vibrating at its second overtone is given by:

f = n × v/2L

where n is the harmonic number. For the second overtone, we have n = 3. Substituting the values, we get:

f = 3 × (166.58 m/s)/(2 × 0.65 m) = 383.33 Hz

We can round off the answer to two significant figures to get:

f ≈ 380 Hz

which is closest to Option D: 383 Hz.

A Heavy Stone of Mass M Is Hung from the Ceiling