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(B) The washer loses contact with the piston when the downward acceleration of the piston exceeds the acceleration due to gravity $(g)$.At the point o

Vedclass · Accepted Answer

$1.9$

Vedclass · Answer

(B) The washer loses contact with the piston when the downward acceleration of the piston exceeds the acceleration due to gravity $(g)$.At the point of losing contact,the normal force $(N)$ becomes $0$.For simple harmonic motion,the maximum downward acceleration is given by $a_{\max} = \omega^2 A$.The condition for losing contact is $a_{\max} = g$,where $g \approx 9.8\, m/s^2$.Given amplitude $A = 7\, cm = 0.07\, m$.Substituting the values: $\omega^2 A = g \Rightarrow \omega = \sqrt{\frac{g}{A}}$.Since $\omega = 2\pi f$,we have $f = \frac{1}{2\pi} \sqrt{\frac{g}{A}}$.$f = \frac{1}{2\pi} \sqrt{\frac{9.8}{0.07}} = \frac{1}{2\pi} \sqrt{140} \approx \frac{11.83}{6.28} \approx 1.88\, Hz$.Rounding to the nearest option,the frequency is $1.9\, Hz$.

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