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KCET · Physics · Waves and Sound

The maximum particle velocity in a wave motion is half the wave velocity. Then the amplitude of the wave is equal to

  1. A \(\frac{\lambda}{4 \pi}\)
  2. B \(\frac{2 \lambda}{\pi}\)
  3. C \(\frac{\lambda}{2 \pi}\)
  4. D \(\lambda\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{\lambda}{4 \pi}\)

Step-by-step Solution

Detailed explanation

For a wave,
\(y=a \sin \frac{2}{\lambda}(v t-x)\)
Differentiating Eq. (i) w.r.t. \(t\), we get
\(\frac{d y}{d t}=\frac{2 \pi v a}{\lambda} \cos \frac{2 \pi}{\lambda}(v t-x)\)
Now, maximum velocity is obtained when
\(\cos \frac{2 \pi}{\lambda}(v t-x)=1\)
\(\therefore v_{\max }=\left(\frac{d y}{d t}\right)_{\max }=\frac{2 \pi v a}{\lambda}\)
but \(v_{\max }=\frac{v}{2}\)
\(\therefore \frac{v}{2}=\frac{2 \pi v a}{\lambda} \Rightarrow a=\frac{\lambda}{4 \pi}\)