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
- A \(\frac{\lambda}{4 \pi}\)
- B \(\frac{2 \lambda}{\pi}\)
- C \(\frac{\lambda}{2 \pi}\)
- D \(\lambda\)
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 \(v_{\max }=\frac{v}{2}\)
\(\therefore \frac{v}{2}=\frac{2 \pi v a}{\lambda} \Rightarrow a=\frac{\lambda}{4 \pi}\)
\(y=a \sin \frac{2}{\lambda}(v t-x)\)
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 \(v_{\max }=\frac{v}{2}\)
\(\therefore \frac{v}{2}=\frac{2 \pi v a}{\lambda} \Rightarrow a=\frac{\lambda}{4 \pi}\)
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