Class 11 Physics - Chapter Waves NCERT Solutions | A travelling harmonic wave on a string i

Welcome to the NCERT Solutions for Class 11th Physics - Chapter Waves. This page offers a step-by-step solution to the specific question from Exercise 1, Question 22: a travelling harmonic wave on a string is describe....
Question 22

A travelling harmonic wave on a string is described by

y(x,t) = 7.5sin [0.0050x + 12t + π/4]

(a) What are the displacement and velocity of oscillation of a point at x = 1 cm, and t = 1 s? Is this velocity equal to the velocity of wave propagation?

 

(b) Locate the points of the string which have the same transverse displacements and velocity as the x = 1 cm point at t = 2 s, 5 s and 11 s.

Answer

(a) The given harmonic wave is:

y(x,t) = 7.5sin [0.0050x + 12t + π/4]

For x = 1 cm and t = 1s,

y = (1, 1) = 7.5sin [0.0050 + 12 + π/4]

.= 7.5sin [12.0050 + π/4]

= 7.5 sinθ

Where, θ = 12.0050 + π/4 =  12.0050  + 3.14 / 4 = 12.79 rad

 =  180 /3.14 x 12.79  =  732.81°

∴ y = (1, 1) = 7.5sin [732.81°]

= 7.5 sin (90 x 8 + 12.81°)

= 7.5 sin (12.81°)

= 7.5 x 0.2217

= 1.6629 ≈ 1.663 cm

The velocity of the oscillation at a given point and time is given as:

v = d/dt y(x,t) = d/dt [7.5sin(0.0050x + 12t +π/4)]

= 7.5 x 12cos (0.0050x + 12t +π/4)

At x = 1 cm and t  =  1s:

v =  y(1,1)  =  90 cos (12.005 + π/4)

= 90cos(732.81°) = 90cos(90 x 8 + 12.81°)

= 90cos(12.81°)

= 90 x 0.975 = 87.75 cm/s

Now, the equation of a propagating wave is given by:

y(x,t) = a sin(kx + wt + ø)

Where,

k = 2π / λ

∴  λ =  2π / k

And ω =  2πv

∴ v  =  ω / 2π

Speed  =  v  =  vλ =  ω / k

Where

ω =  12 rad/s

k  =  0.0050 m-1

∴ v  =  12 /0.0050 = 2400 cm/s

∴ Hence, the velocity of the wave oscillation at x = 1 cm and t = 1 s is not equal to the velocity of the wave propagation.

 

(b) Propagation constant is related to wavelength as:

k  =  2π / λ

λ =  2π / k = 2 x 3.14 / 0.0050

= 1256 cm = 12.56

Therefore, all the points at distances nλ , (n =±1, ±2....and so on)  i.e. ± 12.56 m, ± 25.12 m, … and so on for x = 1 cm, will have the same displacement as the x = 1 cm points at t = 2 s, 5 s, and 11 s.

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