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Waves question

2008 · Shift 2 · Q48
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Waves question

2008 · Shift 2 · Q48

JEE AdvancedPhysicsWavesMCQ+3 / −1
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 12 English
  1. A
    3π50j^{{\sqrt {3\pi } } \over {50}}\widehat j503π​​j​ m/s
  2. B
    −3π50j^- {{\sqrt {3\pi } } \over {50}}\widehat j−503π​​j​ m/s
  3. C
    3π50i^{{\sqrt {3\pi } } \over {50}}\widehat i503π​​i m/s
  4. D
    −3π50i^- {{\sqrt {3\pi } } \over {50}}\widehat i−503π​​i m/s
View written solutionFree

Correct answer: A

  1. Given data
  • Wave speed: v=10 cm/s=0.1 m/sv = 10\ \text{cm/s} = 0.1\ \text{m/s}v=10 cm/s=0.1 m/s
  • Wavelength: λ=0.5 m\lambda = 0.5\ \text{m}λ=0.5 m
  • Amplitude: A=10 cm=0.1 mA = 10\ \text{cm} = 0.1\ \text{m}A=10 cm=0.1 m
  • The wave travels in the positive xxx-direction.

For a transverse wave on a string, particle velocity is along the transverse direction, i.e. along j^\hat jj^​, not along i^\hat ii^. So options C and D are immediately rejected.

  1. Find angular frequency

The frequency is

f=vλ=0.10.5=0.2 Hzf = \frac{v}{\lambda} = \frac{0.1}{0.5} = 0.2\ \text{Hz}f=λv​=0.50.1​=0.2 Hz

Hence angular frequency

ω=2πf=2π(0.2)=2π5 rad/s\omega = 2\pi f = 2\pi(0.2)=\frac{2\pi}{5}\ \text{rad/s}ω=2πf=2π(0.2)=52π​ rad/s
  1. Use SHM relation for particle velocity

Each point of the string performs SHM with amplitude AAA and angular frequency ω\omegaω.

For displacement yyy, transverse velocity is

uy=±ωA2−y2u_y = \pm \omega \sqrt{A^2-y^2}uy​=±ωA2−y2​

Here,

A=0.1 m,y=5 cm=0.05 mA=0.1\ \text{m}, \qquad y=5\ \text{cm}=0.05\ \text{m}A=0.1 m,y=5 cm=0.05 m

So,

A2−y2=(0.1)2−(0.05)2=0.01−0.0025=0.0075=320\sqrt{A^2-y^2} = \sqrt{(0.1)^2-(0.05)^2} = \sqrt{0.01-0.0025} = \sqrt{0.0075} = \frac{\sqrt{3}}{20}A2−y2​=(0.1)2−(0.05)2​=0.01−0.0025​=0.0075​=203​​

Therefore,

∣vy∣=ωA2−y2=2π5⋅320=π350 m/s|v_y| = \omega \sqrt{A^2-y^2} = \frac{2\pi}{5}\cdot \frac{\sqrt{3}}{20} = \frac{\pi\sqrt{3}}{50}\ \text{m/s}∣vy​∣=ωA2−y2​=52π​⋅203​​=50π3​​ m/s
  1. Determine the sign

For a wave moving in the positive xxx-direction,

y(x,t)=f(x−vt)y(x,t)=f(x-vt)y(x,t)=f(x−vt)

which gives

∂y∂t=−v∂y∂x\frac{\partial y}{\partial t}=-v\frac{\partial y}{\partial x}∂t∂y​=−v∂x∂y​

So the transverse velocity has sign opposite to the local slope of the snapshot.

From the given figure, point PPP is at displacement +5 cm+5\ \text{cm}+5 cm on a negative slope portion of the wave. Hence

∂y∂x<0⇒∂y∂t>0\frac{\partial y}{\partial x}<0 \quad \Rightarrow \quad \frac{\partial y}{\partial t}>0∂x∂y​<0⇒∂t∂y​>0

Therefore the point moves upward, i.e. along +j^+\hat j+j^​.

So,

v⃗P=π350 j^ m/s\vec v_P = \frac{\pi\sqrt{3}}{50}\,\hat j\ \text{m/s}vP​=50π3​​j^​ m/s

This matches option A as intended by the problem.

  1. Final answer
v⃗P=π350 j^ m/s\boxed{\vec v_P = \frac{\pi\sqrt{3}}{50}\,\hat j\ \text{m/s}}vP​=50π3​​j^​ m/s​

Hence, the correct option is A.

  1. Comparison with stored answer

The stored correct answer is A, and our derived answer is also A.

Note: The options seem to be written as 3π50\frac{\sqrt{3\pi}}{50}503π​​, but dimensionally and from calculation the result should be π350\frac{\pi\sqrt{3}}{50}50π3​​. Since the stored correct option is A and the direction/sign are correct, I agree with the intended answer choice A.

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