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(A) The condition for the block not to lose contact is $a_{max} \leq g$.$\omega^2 A \leq g$$\omega \leq \sqrt{\frac{g}{A}}$For the minimum time period

Vedclass · Accepted Answer

$0.025$

Vedclass · Answer

(A) The condition for the block not to lose contact is $a_{max} \leq g$.$\omega^2 A \leq g$$\omega \leq \sqrt{\frac{g}{A}}$For the minimum time period $T$,we take $\omega = \sqrt{\frac{g}{A}}$.$T = 2\pi \sqrt{\frac{A}{g}}$Given $A = 1 \ cm = 0.01 \ m$ and $g = 10 \ m/s^2$:$T = 2\pi \sqrt{\frac{0.01}{10}} = 2\pi \sqrt{0.001} \approx 2 	imes 3.14 	imes 0.0316 \approx 0.2 \ s$.The time interval $t = 0$ to $t = 0.05 \ s$ corresponds to $t = 0$ to $t = T/4$.The average potential energy $\langle U angle$ over one-fourth of the time period is given by:$\langle U angle = \frac{1}{T/4} \int_{0}^{T/4} \frac{1}{2} m \omega^2 x^2 \ dt$Since $x = A \sin(\omega t)$,$U = \frac{1}{2} k A^2 \sin^2(\omega t)$.$\langle U angle = \frac{4}{T} \int_{0}^{T/4} \frac{1}{2} m \omega^2 A^2 \sin^2(\omega t) \ dt = \frac{2 m \omega^2 A^2}{T} \int_{0}^{T/4} \sin^2(\omega t) \ dt$Using $\sin^2(	heta) = \frac{1 - \cos(2	heta)}{2}$:$\langle U angle = \frac{m \omega^2 A^2}{T} \int_{0}^{T/4} (1 - \cos(2\omega t)) \ dt = \frac{m \omega^2 A^2}{T} [t - \frac{\sin(2\omega t)}{2\omega}]_{0}^{T/4}$Since $\omega = \frac{2\pi}{T}$,at $t = T/4$,$2\omega t = \pi$:$\langle U angle = \frac{m \omega^2 A^2}{T} [T/4 - 0] = \frac{1}{4} m \omega^2 A^2 = \frac{1}{4} k A^2$Substituting $m = 1 \ kg$ (assumed),$\omega^2 = g/A = 10/0.01 = 1000 \ rad^2/s^2$:$\langle U angle = \frac{1}{4} 	imes 1 	imes 1000 	imes (0.01)^2 = \frac{1}{4} 	imes 1000 	imes 0.0001 = 0.025 \ J$.

For the minimum time period condition,find the average potential energy between $t = 0$ to $t = 0.05 \ s$ (take $g = 10 \ m/s^2$).

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