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PART - 2 CH - 14 Waves — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysicsPART - 2 CH - 14 Waves5 Marks
Question
A transverse harmonic wave on a string is described by
$
y(x, t)=3.0 \sin \left(36 t+0.018 x+\frac{\pi}{4}\right)
$
where $x$ and $y$ are in cm and $t$ in s. The positive direction of $x$ is from left to right.
(a) Is this a travelling wave or a stationary wave?
If it is travelling, what are the speed and direction of its propagation?
(b) What are its amplitude and frequency?
(c) What is the initial phase at the origin?
(d) What is the least distance between two successive crests in the wave?
✓

Answer

The given equation is transverse frquency
$y(x, t)=3.0 \sin \left(36 t+0.018 x+\frac{\pi}{4}\right)$$\quad\ldots$(1)
Standard equation of progressive wave
$y(x, t) =a \sin \left(\frac{2 \pi}{\lambda}(v t-x)+\phi\right) $
$=a \sin \left(\frac{2 \pi}{T} t-\frac{2 \pi}{\lambda} x+\phi\right)$$\quad$$\quad$$\left[\because \frac{1}{T}=\frac{ v }{\lambda}\right]$$\quad\ldots$(2)
(a) By comparing equation (1) with equation (2)
$\frac{2 \pi}{T}=36$
or $\quad \frac{-2 \pi}{\lambda}=0.018$
or$\quad$$\lambda=\frac{-2 \pi}{0.018}\quad\ldots(3)$
$2 \pi n=36$$\quad\ldots$(4)
Multiply equation (3) and equation (4)
$2 \pi n \lambda=\frac{-2 \pi \times 36}{0.018}$
or $\quad n \lambda=\frac{-36 \times 1000}{18}=-2000 cm / s$
Velocity $(v)=-2000 cm / s =-20 m / s$
Here Negative sign represent the wave propagate in right to left.
(b) Given : $\text { speed } =20 m / s $
$a =3.0 cm=3.0 \times 10^{-2} m $
$\frac{2 \pi}{T} =36 $
$\frac{1}{T} =\frac{36}{2 \pi}=\frac{18}{\pi} $
$\text { frequency } n =\frac{18}{3.14}=5.73 Hz$
(c) Phase angle $(\phi)=\frac{\pi}{4}$ Radian
(d) Distance minimum between two consecutive crests wave length $=\lambda$
$=\frac{2 \pi}{0.018} $
$=\frac{2 \times 3.14}{0.018}=348.9 cm $
$\simeq 349 cm $
$\simeq \frac{349}{100} m=3.49 m

$
$\simeq 3.5 m$

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