Maximum amplitude(in cm ) of SHM so block A will not slip on block B , K =100 N / m

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13.Oscillations

hard

Maximum amplitude(in $cm$) of $SHM$ so block A will not slip on block $B , K =100 N / m$

A

$2$

B

$4$

C

$6$

D

$8$

(AIIMS-2019)

Solution

The expression of force is given as,

$F = m \omega^{2} A =\mu mg$

So,

$A =\frac{\mu g }{\omega^{2}}$ And $\omega=\sqrt{\frac{ K }{ m }}$

So,

$\omega=\sqrt{\frac{100}{1.5}}$

Substitute the values.

$A =\frac{(0.4)(9.8)}{\left(\sqrt{\frac{100}{1.5}}\right)^{2}}$

$=6\, cm$

Standard 11

Physics

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