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Question

(C) For a particle executing simple harmonic motion $(SHM)$, the mass $m = 4 	ext{ mg} = 4 	imes 10^{-3} 	ext{ g} = 4 	imes 10^{-6} 	ext{ kg}$.Th

Vedclass · Accepted Answer

$3200 	imes 10^{-6} 	ext{ J m}^{-2}$

Vedclass · Answer

(C) For a particle executing simple harmonic motion $(SHM)$, the mass $m = 4 	ext{ mg} = 4 	imes 10^{-3} 	ext{ g} = 4 	imes 10^{-6} 	ext{ kg}$.The angular frequency is $\omega = 40 	ext{ rad s}^{-1}$.The potential energy is given by $V(x) = a + bx^2$.The restoring force $F$ is given by $F = -\frac{dV}{dx} = -\frac{d}{dx}(a + bx^2) = -2bx$.For $SHM$, the restoring force is also given by $F = -m\omega^2x$.Comparing the two expressions for force, we get $2b = m\omega^2$.Thus, $b = \frac{m\omega^2}{2}$.Substituting the values: $b = \frac{(4 	imes 10^{-6} 	ext{ kg}) 	imes (40 	ext{ rad s}^{-1})^2}{2}$.$b = \frac{4 	imes 10^{-6} 	imes 1600}{2} = 2 	imes 10^{-6} 	imes 1600 = 3200 	imes 10^{-6} 	ext{ J m}^{-2}$.

$A$ particle of mass $4 \text{ mg}$ is executing simple harmonic motion along the $x$-axis with an angular frequency of $40 \text{ rad s}^{-1}$. If the potential energy of the particle is $V(x) = a + bx^2$, where $V(x)$ is in joule and $x$ is in metre, then the value of $b$ is

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