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 of a string 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:$$v=2(0.588\text{ m})(326\text{ Hz})=383\text{ m/s}$$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 $326 \mathrm {~Hz}$