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

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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Sarah 2 months 2021-07-27T22:14:27+00:00 1 Answers 0 views 0

Answers ( )

    0
    2021-07-27T22:15:58+00:00

    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 :/

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