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Waves — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysicsWaves3 Marks
Question
The following equation represents standing wave set up in medium,
$\text{y}=4\cos\frac{\pi\text{x}}{5}\sin40\pi\text{t},$
where x and y are in cm and t in sec. Find out the amplitude and the velocity of the two component waves and calculate the distance between adjacent nodes. What is the velocity of a medium particle at x = 3cm at time $\frac{1}{8}\text{sec}$?
✓

Answer

The given equation of stationary wave is
$\text{y}=4\cos\frac{\pi\text{x}}{3}\sin40\pi\text{t}$
or $\text{y}=2\times2\cos\frac{2\pi\text{x}}{6}\sin\frac{2\text{x}(120)\text{t}}{6}\ \dots{\text{(i)}}$
We know that $\text{y}=2\text{a}\cos\frac{2\pi\text{x}}{\lambda}\sin\frac{2\text{x}v\text{t}}{\lambda}\ \dots(\text{i})$
By comparing tow equations, we get
$\text{a}=2\text{cm}, \lambda=6\text{cm}$ and $v=220\text{cm/ sec}.$
The component waves are:
$\text{y}_1=\text{a}\sin\frac{2\pi}{\lambda}(v\text{t}-\text{x})$ and $\text{y}_2=\text{a}\sin\frac{2\pi}{\lambda}(v\text{t}+\text{x})$
Distance between two adjacent nodes $=\frac{\lambda}{2}=\frac{6}{2}=3\text{cm}.$
Particle velocity $\frac{\text{dy}}{\text{dt}}=4\cos\frac{\pi\text{x}}{3}\cos(40\pi\text{t}).40\pi$
$=160\cos\frac{\pi\text{x}}{3}\cos40\pi\text{t}$
$=160\pi\cos\frac{\pi\text{x}}{3}\cos\Big(40\pi\times \frac{1}{8}\Big)​​=160 \pi$ $[\because \cos\pi=\cos5\pi=-1]$
Hence, particle velocity = 160cm/ sec.

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