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Waves — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsWaves5 Marks
Question
One end of a long string of linear mass density $8.0 \times 10^{–3}kg m^{–1}$ is connected to an electrically driven tuning fork of frequency $256Hz$. The other end passes over a pulley and is tied to a pan containing a mass of $90kg$. 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 $5.0cm$. Write down the transverse displacement y as function of x and t that describes the wave on the string.
✓

Answer

The equation of a travelling wave propagating along the positive y-direction is given by the displacement equation: $\text{y}(\text{x, t})=\text{a}\sin(\text{wt}-\text{kx})\ \dots(\text{i})$ Linear mass density, $\mu=8.0\times10^{-3}\text{kg m}^{-1}$ Frequency of the tuning fork, ν = 256Hz Amplitude of the wave, a = 5.0cm = 0.05m …(ii) Mass of the pan, m = 90kg Tension in the string, $T = mg = 90 \times 9.8 = 882N$ The velocity of the transverse wave v, is given by the relation: $\text{v}=\sqrt{\frac{\text{t}}{\mu}}$
$=\frac{882}{8.0\times10^{-3}}=332\text{m/s}$ Angular Frequency, $\omega=2\pi\text{v}$
$=2\times3.14\times256$
$=1608.5=1.6\times10^3\text{ rad/s}\ \dots(\text{iii})$ Wanelength, $\lambda=\frac{\text{v}}{\text{v}}=\frac{332}{256}\text{m}$
$\therefore$ Propagation constant, $\text{k}=\frac{2\pi}{\lambda}$
$=\frac{2\times3.14}{\frac{332}{256}}=4.84\text{m}^{-1}\ \dots(\text{iv})$Substituting the values from equations (ii), (iii) and (iv) in equatio (i) we get the displacement equation:
$\text{y}(\text{x, t})=0.05\sin(1.6\times10^3\text{t}-4.84\text{x})\text{m}.$

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