Ncert Exercises & Solutions

One end of a long string of linear mass density

8.0 \times 10^{- 3} {kgm}^{- 1}

is connected to an electrically driven tuning fork of frequency 256 Hz. The other end passes over a pulley and is tied to a pan containing a mass of 90 kg. The pulley end absorbs all the incoming energy so that reflected waves at this end have negligible amplitude. At t = 0, the left end (fork end) of the string x = 0 has zero transverse displacement (y = 0) and is moving along positive y-direction. The amplitude of the wave is 5.0 cm. Write down the transverse displacement y as a function of x and t that describes the wave on the string.

The equation of a travelling wave propagating along the positive

x - direction is given by :

y (x, t) = asin (ωt - kx) . . . . (i)

It is given in the question that :

Linear mass density, μ = 8.0 \times 10^{- 3} kg m^{- 1}

Frequency of the tuning fork, ν = 256 Hz

Amplitude of the wave, a = 5.0 cm = 0.05 m . . . . (ii)

Mass of the pan, m = 90 kg

Tension in the string, T = mg = 90 \times 9.8 = 882 N

The velocity of the transverse wave v,

\begin{aligned} v & = \sqrt{\frac{T}{μ}} \\ = \sqrt{\frac{882}{8.0 \times 10^{- 3}}} = 332 m / s \end{aligned}

Angular frequency,

\begin{aligned} ω & = 2 π ν \\ = 2 \times 3.14 \times 256 \\ = 1608.5 = 1.6 \times 10^{3} rad / s . . . (iii) \end{aligned}

Wavelength is given by, λ = \frac{v}{v} = \frac{332}{256} m

∴ Propagation constant is given by,

\begin{aligned} k & = \frac{2 π}{λ} \\ = \frac{2 \times 3.14}{\frac{332}{256}} = 4.84 m^{- 1} . . . (iv) \end{aligned}

Substituting the values from equations (ii), (iii), and (iv) in equation (i),

y (x, t) = 0.05 \sin (1.6 \times 10^{3} t - 4.84 x) m

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