Question

Two blocks A & B in which block B is firmly fixed on the ground & block A is firmly fixed on block B . Whether this system oscillates when you applied a force on block A ? What is the time period of oscillation ?

i) $2\pi {\sqrt{M{\eta}L}}$
ii) $2\pi {\sqrt{\dfrac{M{\eta}}{L}}}$
iii) $2\pi {\sqrt{\dfrac{M{L}}{\eta}}}$
iv) $2\pi {\sqrt{\dfrac{M}{{\eta}L}}}$



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Answer 1

Yes, the system will have oscillation if a force F is applied on the block A.

Let's suppose the system undergoes a very little displacement during oscillation ,let's say $\Delta x$

We have shearing stress/strain($\eta$) :

$\eta = \frac{FL} {A \Delta x} \: \: \: \: \: [ \Delta x= magnitude~ of ~displacement.] \\\\ F= \frac{ \eta A} {L} \Delta x - - - - [i]$

As nothing is provided in the question, I took $\eta$ is shearing stress/strain

By definition of SHM, we have acceleration, $a = \omega^2 \Delta x - - - - [ii]$ where $\omega^2$ is a positive constant.

Now, we have,

$F= ma \\\\ a= \frac{F} {m} - - - - [iii]$

Equating [ii] and [iii], we have

$\frac{F}{m} = \omega^2 \Delta x \\\\ F = m \omega^2 \Delta x \\\\ F= k \Delta x - - - - [iv]$

We also know that, time period of oscillation is :

$T= 2 \pi \sqrt{\frac{M}{k}} - - - - [v]$

Comparing [i] and [iv], we have:

$k \Delta x = \frac{ \eta A} {L} \Delta x \\\\ k \cancel{\Delta x} = \frac{ \eta A} {L} \cancel{ \Delta x} \\\\ k= \frac{ \eta A} {L}$

Put this value of k in [v],

$T= 2 \pi \sqrt{\frac{M}{ \frac{ \eta A} {L} }} \\\\ T= 2\pi {\sqrt{\dfrac{M{L}}{A \eta}}}$

we take Area = 1, as its not mentioned in the question, so the required time period is:

$T= 2\pi {\sqrt{\dfrac{M{L}}{ \eta}}}$

Option (iii) is correct, not sure tho :/