Question

What will happen to fundamental frequency of a wave in a stretched string when tension of the string increases 4 times and its length reduced to half of the?

Answer 1

Answer: The new frequency will be 4 times the original frequency.

Explanation:

Frequency of a string under tension is given by the following equation

f = $\frac{p}{2l}$$\sqrt{\frac{T}{m} }$

Which means frequency at which the string vibrates is proportional to the $\frac{1}{2}$  the tension in the string.

and it is inversely proportional to the length of the string.

Let $T_1$ is the new tension , which is 4 times the original tension.

$T_1$ = 4T  .............equation 1

and $l_1$ is the new length which is half the original length.

$l_1$  = $\frac{l}{2}$   ..............equation 2

and let $f_1$ be the new frequency,

We have

$\frac{f_1}{f}$ = $\frac{l}{l_1}$$\sqrt{\frac{T_1}{T} }$

Substituting values from equation 1 and 2, we have,

$\frac{f_1}{f}$  = $\frac{2l}{l}$$\sqrt{\frac{4T}{T} }$

$\frac{f_1}{f}$ = 4

∴ $f_1$ = 4f

Hence, The new frequency will be 4 times the original frequency.

Answer 2

Given:

Tension in a stretched string increases 4 times and length reduced to half .

To Find:

Change in Fundamental frequency of the wave in  string.

Solution:

Fundamental frequency is defined as the  lowest frequency which is produced by the oscillation of the whole of an object, as distinct from the harmonics of higher frequency.

$f_{n} = \frac{nv}{2L}\\ v = \sqrt{\frac{T}{\mu} }$

Where T is the tension in string and $\mu$ is mass per unit length of string.

L is the length of string.

From the above formula w can formulate that

$f = k\frac{\sqrt{T} }{L}$

where K is constant of proportionality.

T = 4 T          L = .5 L

$f = k \frac{\sqrt{4T} }{.5L} \\f = 4k \frac{T}{L}$

The fundamental frequency becomes 4 times of the original fundamental frequency.