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QJEEAdv08Phfe07: JEE Advanced 2008 Physics - Class 11 JEEAdv08Phfe07 Solution: A transverse sinusoidal wave moves along a string in the…

A transverse sinusoidal wave moves along a string in the positive x-direction at a speed of 10 cm/s. The wavelength of the waves is 0.5 m and its amplitude is 10 cm. At a particular time t, the snap-shot of the wave is shown in figure. The velocity of point P when its displacement is 5 cm is :

IIT-JEE 2008 Paper 2 Offline Physics - Waves Question 11 English

Multiple Choice Options

  • A 3π50j^{{\sqrt {3\pi } } \over {50}}\widehat j m/s
  • B −3π50j^- {{\sqrt {3\pi } } \over {50}}\widehat j m/s
  • C 3π50i^{{\sqrt {3\pi } } \over {50}}\widehat i m/s
  • D −3π50i^- {{\sqrt {3\pi } } \over {50}}\widehat i m/s

Step-by-Step Verified Solution

Velocity of point p=dy/dtp=dy/dt

= −- (velocity of wave ×\times dy/dxdy/dx )

Here, dy/dx=dy/dx= negative ( −- ve). Therefore, velocity at point P is positive and is along X-axis only.

Equation of a wave moving in positive X-axis is given as,

y=Asin⁡(wt−ϕ)y = A\sin (wt - \phi ) .... (i)

or, VP=Awcos⁡(wt−ϕ){V_P} = Aw\cos (wt - \phi ) .... (ii)

Here, y′=5y' = 5 cm, A=10A = 10 cm

∴\therefore 5=10sin⁡(wt−ϕ)⇒wt−ϕ=30∘5 = 10\sin (wt - \phi ) \Rightarrow wt - \phi = 30^\circ .... (iii)

Put equation (iii) in equation (ii),

VP=0.10×wcos⁡30∘{V_P} = 0.10 \times w\cos 30^\circ

Now, V=VλV = {V_\lambda } , ∴\therefore v=V/λ=0.100.5=0.2v = V/\lambda = {{0.10} \over {0.5}} = 0.2

∴\therefore w=2πv=2π×0.2=0.4πw = 2\pi v = 2\pi \times 0.2 = 0.4\pi

⇒VP=0.1×0.4π×32=350π j^\Rightarrow {V_P} = 0.1 \times 0.4\pi \times {{\sqrt 3 } \over 2} = {{\sqrt 3 } \over {50}}\pi \,\widehat j m/s

It is in positive Y-direction.

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