Question

A string that is fixed at both ends has a length of 1.81 m. When the string vibrates at a frequency of 84.3 Hz, a standing wave with five loops is formed. (a) What is the wavelength of the waves that travel on the string? (b) What is the speed of the waves? (c) What is the fundamental frequency of the string?

Answer 1

Answer:

(a) λ= 0.603m

(b) v= 50.8329$m/s$

(c) $f_{o}=14.042Hz$

Explanation:

(a)The string is an open pipe and five loops indicates that the wave forms three complete cycle in the string.

A circle represents 1 wavelength. The loops occupy the whole length of the pipe, hence;

3λ = 1.81m

 λ= $\frac{1.81}{3}$

 λ= 0.603m

(b) speed(v) = fλ

               v= 84.3 * 0.603

               v= 50.8329$m/s$

(c) The string represents an open pipe.for an open pipe, the fundamental frequency is $f_{o}=\frac{v}{2L}$

$f_{o}=14.042Hz$