Question

For a string stretched between two supports, two successive standing-wave frequencies are 535 Hz and 642 Hz. There are other standing-wave frequencies lower than 535 Hz and higher than 642 Hz.

Answer 1

Answer:

The length of the string is 1.73 m.

Explanation:

Given that,

Successive frequency f= 535 Hz

Successive frequency f' =642 Hz

Suppose, if the speed of transverse waves on the string is 372 m/s , what is the length of the string? Assume that the mass of the wire is small enough for its effect on the tension in the wire to be neglected.

The nth harmonic is

$f_{n}=\dfrac{nv}{2L}$....(I)

The (n+1)th harmonic is

$f_{n+1}=\dfrac{(n+1)v}{2L}$....(II)

We need to calculate the length of the string

Subtract equation (I)  from equation(I)

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

$f_{n+1}-f_{n}=\dfrac{v}{2L}$

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

Put the value into the formula

$L=\dfrac{372}{2(642-535)}$

$L=1.73\ m$

Hence, The length of the string is 1.73 m.

Answer 2

Answer:

2.064 m

Explanation:

Supposing we have to find the length of the string L also considering the speed of the wave to be 384 m/s.

We know that when a string of a definite length L rigidly held at both ends, there is a standing wave formed where its possible frequencies are given by:

$f_n =n\frac{\nu}{2L}$

It is given that 535 Hz and 642 Hz are two adjacent frequencies, so if 535 Hz is f_n  then 642 Hz will be f_n+1

therefore, we can write

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

therefore,

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

plugging values

[tex]L =\frac{384}{2(642-535)}[\tex]

= 2.064 m