Wave Motion Ques 9
- A harmonically moving transverse wave on a string has a maximum particle velocity and acceleration of $3 m / s$ and $90 m / s^{2}$ respectively. Velocity of the wave is $20 m / s$. Find the waveform.
(2005, 2M)
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Answer:
Correct Answer: 9.$y=(0.1 \mathrm{~m}) \sin \left[(30 \mathrm{rad} / \mathrm{s}) t \pm\left(1.5 \mathrm{~m}^{-1}\right) x+\phi\right]$
Solution:
Formula:
Terms Related to Wave Motion (For 1-D Progressive Sine Wave):
- Maximum particle velocity,
$$ \omega A=3 m / s \cdots(i) $$
Maximum particle acceleration,
$$ \omega^{2} A=90 m / s^{2} \cdots(ii) $$
Velocity of wave, $$\frac{\omega}{k}=20 m / s\cdots(iii)$$
From Eqs. (i), (ii) and (iii), we get
$$ \omega=30 rad / s \Rightarrow A=0.1 m \quad \text { and } \quad k=1.5 m^{-1} $$
$\therefore$ Equation of waveform should be
$$ \begin{aligned} & y=A \sin (\omega t+k x+\phi) \\ & y=(0.1 m) \sin \left[(30 rad / s) t \pm\left(1.5 m^{-1}\right) x+\phi\right] \end{aligned} $$
BETA
