Question

A standing wave is produced on a string fixed at one end and free at other. The length of string must be an integral multiple of lambda

Answer 1

Answer:

For the given end conditions, standing waves are produced if ν = (n+½)v/2L →v/ = (n+½)v/2L →L = (n+½)/2 =(2n+1)*/4, where n = 0, 1, 2, .... Thus the length of the string is an integral multiple of /4.Read more on Sarthaks.com - https://www.sarthaks.com/40559/a-standing-wave-is-produced-on-a-string-clamped-at-one-end-and-free-at-the-other

Answer 2

The length of the string is an integral multiple of $\dfrac{1}{4}$

Explanation:

We know that,

The frequency of the wave is

$f=\dfrac{v}{\lambda}$...(I)

According to given condition,

We need to calculate the integral multiple of λ

Using formula of frequency

$f=(n+\dfrac{1}{2})\times\dfrac{v}{2L}$

Put the value of frequency from equation (I)

$\dfrac{v}{\lambda}=(n+\dfrac{1}{2})\times\dfrac{v}{2L}$

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

If , n = 0, 1, 2.....

Put the value of n in equation (II)

The length will be

$L_{0}=\dfrac{\lambda}{4}$

$L_{1}=\dfrac{3\lambda}{4}$

$L_{2}=\dfrac{5\lambda}{4}$

Hence, The length of the string is an integral multiple of $\dfrac{1}{4}$

Learn more :

Topic : standing wave  

https://brainly.in/question/3565219