alleryFor a sinusoidal mechanical wave $y=A\sin(\omega t\pm kx)$, the maximum particle speed is
A$A/k$
B$\omega/k$
C$A\omega$
D$Ak$
Answer & Solution
Correct answer: C. $A\omega$
Particle velocity is $v_p=\dfrac{dy}{dt}=A\omega\cos(\omega t\pm kx)$.
Since the maximum value of cosine is 1, the maximum particle speed is $(v_p)_{\max}=A\omega$.
**Therefore option C is correct.**
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