Ncert Exercises & Solutions

The transverse displacement of a string (clamped at its both ends) is given by $y(x,t)=0.6 sin\left ( \frac{2\pi }{3} x\right )cos (120 \pi t)$ where x and y are in m and t in s. The length of the string is 1.5m and its mass is $3.0\times 10^{-2} kg$.
Answer the following:
(a) Does the function represent a traveling wave or a stationary wave?
(b) Interpret the wave as a superposition of two waves traveling in opposite directions. What is the wavelength, frequency, and speed of each wave?
(c) Determine the tension in the string.

The general equation representing a stationary wave:

y (x, t) = 2asinkxcos ωt
This equation is similar to the equation:
$y (x, t) = 0.06 \sin (\frac{2 π}{3} x) \cos (120 πt)$
Hence, the given function represents a stationary wave.
A wave travelling along the positive x-direction is:
$y_{1} = a \sin (ωt - kx)$

The wave travelling along the negative x-direction is:
$y_{2} = a \sin (ωt + kx)$

The superposition of these two waves yields:

$\begin{array}{l} y = y_{1} + y_{2} = asin (ωt - kx) - asin (ωt + kx) \\ = asin (ωt) \cos (kx) - asin (kx) \cos (ωt) - asin (wt) \cos (kx) - asin (kx) \cos (wt) \\ = - 2 asin (kx) \cos (ωt) \\ = - 2 asin (\frac{2 π}{λ} x) \cos (2 πνt) . . . (i) \end{array}$

The transverse displacement of the string is:

$y (x, t) = 0.06 \sin (\frac{2 π}{3} x) \cos (120 πt) . . . (ii)$

Comparing equations (i) and (ii),

$Wavelength, λ = 3 m$
$Frequency, ν = 60 Hz$
$Wave speed, v = νλ = 60 \times 3 = 180 m / s$
$The velocity of a transverse wave travelling in a string :$
$v = \sqrt{\frac{T}{μ}} . . . (i)$
$Velocity of the transverse wave, v = 180 m / s$
$Mass of the string, m = 3.0 \times 10^{2} kg$
$Length of the string, l = 1.5 m$
$Mass per unit length of the string, μ = \frac{m}{l} = \frac{3.0}{1.5} \times 10^{- 2} = 2 \times 10^{- 2} kg m^{- 1}$
$Tension in the the string,$
$T = v^{2} μ$
$= {(180)}^{2} \times 2 \times 10^{- 2}$
$= 648 N$

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