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) [tex] 2\pi {\sqrt{M{\eta}L}} [/tex]
ii) [tex] 2\pi {\sqrt{\dfrac{M{\eta}}{L}}} [/tex]
iii) [tex] 2\pi {\sqrt{\dfrac{M{L}}{\eta}}} [/tex]
iv) [tex] 2\pi {\sqrt{\dfrac{M}{{\eta}L}}} [/tex]
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 [tex] \Delta x [/tex]
We have shearing stress/strain([tex] \eta[/tex]) :
[tex] \eta = \frac{FL} {A \Delta x} \: \: \: \: \: [ \Delta x= magnitude~ of ~displacement.] \\\\ F= \frac{ \eta A} {L} \Delta x – – – – [i] [/tex]
As nothing is provided in the question, I took [tex] \eta [/tex] is shearing stress/strain
By definition of SHM, we have acceleration, [tex] a = \omega^2 \Delta x – – – – [ii] [/tex] where [tex] \omega^2 [/tex] is a positive constant.
Now, we have,
[tex] F= ma \\\\ a= \frac{F} {m} – – – – [iii] [/tex]
Equating [ii] and [iii], we have
[tex] \frac{F}{m} = \omega^2 \Delta x \\\\ F = m \omega^2 \Delta x \\\\ F= k \Delta x – – – – [iv] [/tex]
We also know that, time period of oscillation is :
[tex] T= 2 \pi \sqrt{\frac{M}{k}} – – – – [v] [/tex]
Comparing [i] and [iv], we have:
[tex] 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} [/tex]
Put this value of k in [v],
[tex] T= 2 \pi \sqrt{\frac{M}{ \frac{ \eta A} {L} }} \\\\ T= 2\pi {\sqrt{\dfrac{M{L}}{A \eta}}} [/tex]
we take Area = 1, as its not mentioned in the question, so the required time period is:
[tex] T= 2\pi {\sqrt{\dfrac{M{L}}{ \eta}}} [/tex]
Option (iii) is correct, not sure tho :/