Standing waves are set up on two strings fixed at each end. The two strings have

Question

Standing waves are set up on two strings fixed at each end. The two strings have the same tension, but they differ in length by 0.57 cm. The waves on the shorter string propagate with a speed of 41.8 m/s, and the fundamental frequency of the shorter string is 225 Hz. What is the shortest tube you could use to produce a standing wave in air of the same fundamental frequency as the shorter string (225 Hz)? Would it be an open-open (or closed-closed) tube or an open-closed tube? Assume that the speed of sound in air is 343 m/s.

Explanation / Answer

For a wave on the string, the speed v, tension T and linear density (mass per unit length, u) are related by

v = (T/u).

T and u are the same for both pieces of string, so the wavespeed must be the same for both pieces of string.

Also for any wave, v = f, speed = frequency times wavelength. So for the shorter string, = v/f
= (41.8 m/s)/(225 Hz) = 0.18577 m.

And for the fundamental wave, the length of string is half a wavelength, or 0.18577 m/2 = 9.28889 cm.

The longer piece of string is 0.57 cm longer than the shorter piece, so its length is 9.288889 cm, and the fundamental wavelength is double this, or 18.57777 cm = 0.1857777 m.

We have already ascertained that the speed of the waves is the same in both waves, so v = 41.8 m/s on the longer string too. And f = v/ = (41.8 m/s)/(0.185777 m) = 225 Hz.

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.