Question

A standing wave is produced on a string clamped at one end and free at the other. The length of the string
(a) must be an integral multiple of λ/4
(b) must be an integral multiple of λ/2
(c) must be an integral multiple of λ
(d) must be an integral multiple of λ/2.

Answer 1

Length of String = Integral Multiple of 1/4

Explanation:

For the given end conditions, standing waves are produced if -

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

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

$L = \frac {n + \frac {1}{2}}{2} = (2n + 1) \times \frac {1}{4}$ ....where n =0,1,2,3,,,,

Thus the length of the string is an integral multiple of 1/4.

Answer 2

(a)The length of the string is an integral multiple of λ/4.

Explanation:

The fundamental frequency of a standing wave which is produced on a string clamped at one end and free at the other is given as

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

Here, v is the velocity and L is the length of the string and n is integer.

We know the velocity, frequency and wavelength relation is given as

$v=\nu\lambda$

so above from formula,

$\nu=(n+\frac{1}{2})\frac{\nu\lambda}{2L}$

or,

$L=(2n+1)\frac{\lambda}{4}$

For n = 0, 1, 2, 3......

$L=\frac{\lambda}{4},\frac{3\lambda}{4},\frac{5\lambda}{4}.........$

Thus, the length of the string is an integral multiple of λ/4.

#Learn More: standing wave.

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