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:* $f_1$ is the lowest standing wave frequency* $v$ is the wave speed* $L$ is the length of the stringSolving for $v$, we get:$$v = 2f_1L$$Substituting the given values, we get:$$v = 2(326 \mathrm{~Hz})(0.588 \mathrm{~m}) = 383 \mathrm{~m/s}$$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 $326 \mathrm {~Hz}$