JEE Main 2022 · 26 Jul · Shift 2 · Physics Q43

← JEE Main 2022 · 26 Jul · Shift 2

PhysicsSingle correct+4 / -1Question 43 of 90

A transverse wave is represented by $y=2\sin(\omega t-kx)$ cm. The value of wavelength (in cm) for which the wave velocity becomes equal to the maximum particle velocity, will be :

4\pi

2\pi

\pi

Answer and worked solution

Official answer: A

Equate $\frac\omega k=A\omega$ with $A=2$ cm: $k=\frac12$ , so $\lambda=\frac{2\pi}k=4\pi$ cm.

(A)

Topics:Wave equation and progressive waves

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JEE Main 2022 · 26 Jul · Shift 2, Physics question 43