Standing waves on a string: 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 speed of the pulse in the wire can be calculated using the time it takes for the pulse to travel down and back, which is $7.84 \, \text{ms} = 0.00784 \, \text{s}$. Since the length of the wire is $65.0 \, \text{cm} = 0.65 \, \text{m}$, the total distance traveled by the pulse is twice this length, or $2 \times 0.65 \, \text{m} = 1.3 \, \text{m}$. The speed of the pulse ($v$) is then $v = \frac{distance}{time} = \frac{1.3 \, \text{m}}{0.00784 \, \text{s}} \approx 165.8 \, \text{m/s}$.For a string fixed at both ends, the fundamental frequency ($f_1$) is given by $f_1 = \frac{v}{2L}$, where $L$ is the length of the string. The frequencies of the overtones are multiples of the fundamental frequency. The second overtone corresponds to the third harmonic ($n = 3$), so its frequency ($f$) is given by $f = n \times f_1 = 3 \times \frac{v}{2L}$.Substituting the given values, $f = 3 \times \frac{165.8 \, \text{m/s}}{2 \times 0.65 \, \text{m}} \approx 382.5 \, \text{Hz}$, which rounds to $383 \, \text{Hz}$, making option D the correct answer.

Standing Waves on a String: a Heavy Stone of Mass