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Question

A sinusoidal wave travelling in the positive direction on a stretched string has an amplitude 20 cm, wavelength 1.0 m and wave velocity 5 m/s. At x=0 and t=0 , it is given that y=0 and yt<0. Find the wave function y(x,t)

A
y(x.t)=0.02sin[(2πx)+(10πt)] m
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B
y(x.t)=0.02cos[(10πx)+(2πt)] m
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C
y(x.t)=0.02sin[(2πx)(10πt)] m
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D
y(x.t)=0.02sin[(πx)+(5πt)] m
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Solution

The correct option is C y(x.t)=0.02sin[(2πx)(10πt)] m
General equation of wave travelling in positive x direction
y(x,t)=Asin(kxωt+ϕ)
Given that
Amplitude (A)=20 cm or 0.02 m
wavelength (λ)=1 m
Angular wave number (k)=2πλ=2π m1
Angular frequency ω=vk=10π rad/s
So, y(x,t)=0.02sin[2πx10πt+ϕ] ........(1)

From the data given in the question ,

for x=0,t=0 y=0 & yt<0

So, for y=0, we get
0.02sinϕ=0sinϕ=nπ ....(2) Where n=0,1,2,3.....

For , yt<0
0.2πcosϕ<00.2cosϕ>0cosϕ>0 .......(3)

From (2) and (3) we get,
ϕ=2nπ where n=0,2,4,6....
Now we know that
sin(2π+θ)=sinθ
Therefore,
y(x,t)=0.02sin[(2πx)(10πt)] m

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