Question

der a string stretched between two supports. The speed of the propagation of transverse waves is 85 m/s, and two successive standing-wave frequencies are 510 Hz and 680 Hz.a) What is the the order of these two harmonics (first, second, etc)?b) What is the length of the string?

Answer 1

Answer:

a) 3, 4

b)0.25 m

Explanation:

Given frequency f_1= 510 Hz

and frequency f_2= 680 Hz

v= 85 m/s

$f_n= \frac{nv}{2L}$

and

$f_{n+1}= \frac{(n+1)v}{2L}$

therefore,

$f_{n+1}-f_n= \frac{(n+1)v}{2L}-\frac{nv}{2L}$

= v/2L

$L= \frac{v}{2(f_{n+1}-f_1)}$

$L= \frac{85}{2(680-510)}$

Therefore, length of the string = 0.25 m

510 = 85n/2×0.25

n= 3

therefore order of two harmonics 3,4 .