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?
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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?
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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