A block of mass M rests on a piston executing | vedclass

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(A) For a particle executing $S.H.M.$, the acceleration is given by $a = \omega^2 x$. The maximum acceleration is $a_{max} = \omega^2 A$.For the block

Vedclass · Accepted Answer

$0.25 \,m$

Vedclass · Answer

(A) For a particle executing $S.H.M.$, the acceleration is given by $a = \omega^2 x$. The maximum acceleration is $a_{max} = \omega^2 A$.For the block to separate from the piston, the downward acceleration of the piston must exceed the acceleration due to gravity, $g$.Thus, the condition for separation is $a_{max} \geq g$.Given the period $T = 1 \,s$, we calculate the angular frequency $\omega = \frac{2\pi}{T} = 2\pi \,rad/s$.Substituting the condition $a_{max} = g$:$\omega^2 A = g$$(2\pi)^2 A = 10$$4\pi^2 A = 10$Given $\pi^2 = 10$, we have $4(10) A = 10$.$40 A = 10$$A = \frac{10}{40} = 0.25 \,m$.

$A$ block of mass $M$ rests on a piston executing $S.H.M.$ with a period of $1 \,s$. The amplitude of oscillations, such that the mass is separated from the piston, is (acceleration due to gravity, $g=10 \,m/s^2$, $\pi^2=10$)

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