MCQ
In a simple harmonic wave, minimum distance between particles in same phase always having same speed, is ..........
- ✓$\lambda / 4$
- B$\lambda / 3$
- C$\lambda / 2$
- D$\lambda$
✓
Answer
Correct option: A.
$\lambda / 4$
a
(a)
(a)
$y=A \sin (\omega t-k x)$
$\frac{d y}{d t}=A \omega \cos (\omega t-k x)$
Take $x=0$ and $x=x_1$
$v_1=A \omega \cos (\omega t)$
$v_2=A \omega \cos (\omega t+k x)$
$\left|v_1\right|=\left|v_2\right|$
$|\cos (\omega t)|=|\cos (\omega t+k x)|$
$\therefore k x=\frac{\pi}{2}$
or $x=\frac{\lambda}{4}$
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