A 0.588-m string is tightly clamped at both ends. If the lowest standing wave frequency of the string is $326 \mathrm {~Hz}$ how fast do waves travel on this string?
A) 383 m/s
B) 475 m/s
C) 582 m/s
D) 724 m/s

Question

Quizplus · Accepted Answer

The lowest standing wave frequency is given by:$$f_1 = \frac{v}{2L}$$where $v$ is the wave speed and $L$ is the string length. Solving for $v$, we get:$$v = 2Lf_1$$Plugging in the given values, we get:$$v = 2(0.588 \mathrm{~m})(326 \mathrm{~Hz}) = 383 \mathrm{~m/s}$$Therefore, the correct answer is A.

A 0588-M String Is Tightly Clamped at Both Ends 326 Hz326 \mathrm {~Hz}326 Hz

A 0588-M String Is Tightly Clamped at Both Ends $326 \mathrm {~Hz}$