Question

A string 2m in length sustains a standing wave with points of string at which displacement amplitude is 2root2mm being separated by 15cm and 5cm alternatively. Then the amplitude of the antinode will be?​

Answer 1

Answer ⤵️

From figure, points A,B,C,D,E and F have equal displacement amplitudes.

x

E

−x

A

=λ=4×15cm=60cm

Thus n=

60/2

120

=4

Thus it corresponds to the 4th harmonic.

Distance of node from point A=7.5cm.

Amplitude of stationary wave can be written as a=AsinKx

Here, a=

2

mm, K=

λ

2π

=

60

2π

and x=7.5cm

⟹

2

=Asin(

60

2π

×7.5)=Asin

4

π

=

2

A

⟹A=2mm