A transverse wave is described by the equatio | vedclass

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(B) The equation of the wave is $y = y_0 \sin 2\pi \left( ft - \frac{x}{\lambda} \right)$.The particle velocity $v_p$ is given by the derivative of di

Vedclass · Accepted Answer

$\lambda = \frac{\pi y_0}{2}$

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(B) The equation of the wave is $y = y_0 \sin 2\pi \left( ft - \frac{x}{\lambda} \right)$.The particle velocity $v_p$ is given by the derivative of displacement with respect to time: $v_p = \frac{\partial y}{\partial t} = y_0 (2\pi f) \cos 2\pi \left( ft - \frac{x}{\lambda} \right)$.The maximum particle velocity is $V_{\max} = 2\pi f y_0$.The wave velocity $V$ is given by $V = f\lambda$.According to the problem,$V_{\max} = 4V$.Substituting the expressions,we get: $2\pi f y_0 = 4(f\lambda)$.Dividing both sides by $f$,we get: $2\pi y_0 = 4\lambda$.Therefore,$\lambda = \frac{2\pi y_0}{4} = \frac{\pi y_0}{2}$.

List-$I$	List-$II$
$(A)$ Magnetic field intensity	$(i)$ $Wb$
$(B)$ Magnetic flux	(ii) $Wb \cdot m^{-2}$
$(C)$ Magnetic pole strength	(iii) $A \cdot m$
$(D)$ Magnetic induction	(iv) $A \cdot m^{-1}$

$A$ transverse wave is described by the equation $y = y_0 \sin 2\pi \left( ft - \frac{x}{\lambda} \right)$. The maximum particle velocity is equal to four times the wave velocity if

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