Answer to Question #195536 in Mechanics | Relativity for Mohammad Balhas

Question

A standing wave with wavelength λ = 1.2 m and frequency f = 50 Hz is generated on a stretched cord. For an element of the cord at x = 0.5 m, the maximum transverse velocity is v(y,max) = 2π m/s. The amplitude A of each of the individual waves producing the standing wave is:

0.0125 m

0.03 m

0.02 m

0.01 m

0.025 m

Answer 1

Gives

$\lambda=1.2m$

$f=50Hz$

$x=0.5m$

$V_{max}=2\pi$ m/sec

We know that

Individual wave equation

$y=Asin(wt-kx)\rightarrow(1)$

eqution differenciate with respect time

$V=Awcos(wt-kx)$

Phase $(wt-kx)=0°$

Then $cos(wt-kx)=cos0°=1$

$V=Aw\rightarrow(2)$

$2\pi=Aw$

$2\pi=A2\pi\times50$

$A=\frac{1}{50}=0.02m$

Amplitude of each standing wave(A)=0.02m

Option (c) is correct option