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

Question

Quizplus · Accepted Answer

The fundamental frequency (lowest standing wave frequency) of a string clamped at both ends is given by f = v / (2L), where v is the wave speed and L is the length of the string. Solving for v gives v = 2Lf. Substituting L = 0.588 m and f = 326 Hz, we get v = 2 * 0.588 m * 326 Hz = 383 m/s.

A 0588-M String Is Tightly Clamped at Both Ends