In a simple harmonic wave,the minimum distanc | vedclass

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(A) The displacement of a particle in a simple harmonic wave is given by $y = A \sin(\omega t - kx)$.The velocity of the particle is $v = \frac{dy}{dt

Vedclass · Accepted Answer

$\lambda / 4$

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(A) The displacement of a particle in a simple harmonic wave is given by $y = A \sin(\omega t - kx)$.The velocity of the particle is $v = \frac{dy}{dt} = A\omega \cos(\omega t - kx)$.Consider two particles at positions $x_1$ and $x_2$. Their velocities are $v_1 = A\omega \cos(\omega t - kx_1)$ and $v_2 = A\omega \cos(\omega t - kx_2)$.For the speeds to be equal,$|v_1| = |v_2|$,which implies $|\cos(\omega t - kx_1)| = |\cos(\omega t - kx_2)|$.This condition is satisfied when the arguments differ by $\pi/2$ or $\pi$. The minimum distance corresponds to a phase difference of $\pi/2$.Thus,$k(x_2 - x_1) = \frac{2\pi}{\lambda} \Delta x = \frac{\pi}{2}$.Solving for $\Delta x$,we get $\Delta x = \frac{\lambda}{4}$.

In a simple harmonic wave,the minimum distance between particles in the same phase always having the same speed is ..........

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