Question

A wave travelling along the positive x−direction having maximum displacement along y−direction as 1 m, wavelength 2π m and frequency of 1/π Hz is represented by:

- y=sin(2πx−2πt)
- y=sin(10πx−20πt)
- y=sin(x−2t)
- y=sin(2πx+2πt)

Solution

The correct option is **C** y=sin(x−2t)

Maximum vertical( y−direction) displacement of particle on string is the amplitude of wave

⇒A=1 m

Wavelength(λ)=2π m,

frequency (f)=1π Hz

Let the equation of wave be,

y=Asin(kx−ωt+ϕ)

where k=2πλ and ω=2πf

⇒y=1×sin(2π2πx−2ππt+ϕ)

⇒y=sin(x−2t+ϕ)

Now if at time t=0, particle on string at x=0 starts motion from its mean position i.e y=0 then phase constant ϕ=0.

Hence the wave equation will reduce to,

y=sin(x−2t)

∴Option (a) is correct.

Maximum vertical( y−direction) displacement of particle on string is the amplitude of wave

⇒A=1 m

Wavelength(λ)=2π m,

frequency (f)=1π Hz

Let the equation of wave be,

y=Asin(kx−ωt+ϕ)

where k=2πλ and ω=2πf

⇒y=1×sin(2π2πx−2ππt+ϕ)

⇒y=sin(x−2t+ϕ)

Now if at time t=0, particle on string at x=0 starts motion from its mean position i.e y=0 then phase constant ϕ=0.

Hence the wave equation will reduce to,

y=sin(x−2t)

∴Option (a) is correct.

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