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 $256\; Hz$. 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 function of $x$ and $t$ that describes the wave on the string.
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 $256\; Hz$. 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 function of $x$ and $t$ that describes the wave on the string.
The equation of a travelling wave propagating along the positive $y$ -direction is given by the
displacement equation: $y(x, t)=a \sin (w t-k x) \ldots(i)$
$\mu=8.0 \times 10^{-3}\, kg\, m ^{-1}$
Linear mass density,
Frequency of the tuning fork, $v=256\, Hz$
Amplitude of the wave, $a=5.0\, cm =0.05\, m \ldots (ii)$
Mass of the pan, $m=90 \,kg$
Tension in the string, $T=m g=90 \times 9.8=882\, N$
The velocity of the transverse wave $v$, is given by the relation
$v=\sqrt{\frac{T}{\mu}}$
$=\sqrt{\frac{882}{8.0 \times 10^{-3}}}=332 \,m / s$
Angular frequency, $\omega=2 \pi v$ $=2 \times 3.14 \times 256$
$=1608.5=1.6 \times 10^{3}\, rad / s\ldots(iii)$
Wavelength, $\lambda=\frac{v}{v}=\frac{332}{256} \,m$
$\therefore$ Propagation constant, $k=\frac{2 \pi}{\lambda}$
$=\frac{2 \times 3.14}{\frac{332}{256}}=4.84 \,m ^{-1}\ldots(i v)$
Substituting the values from equations $ (ii), (iii)$, and $(iv)$ in equation ($i$), we get the displacement equation:
$y(x, t)=0.05 \sin \left(1.6 \times 10^{3} t-4.84 x\right)\, m$
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