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Passage IV) Angular frequency in SHM is ...

Passage IV) Angular frequency in SHM is given by `omega=sqrt(k/m)`. Maximum acceleration in SHM is `omega^(2)` A and maximum value of friction between two bodies in contact is `muN`, where N is the normal reaction between the bodies.

In the figure shown, what can be the maximum amplitude of the system so that there is no slipping between any of hte blocks?

A

`2/7 m`

B

`3/4 m`

C

`4/9 m`

D

`10/3 m`

Text Solution

Verified by Experts

The correct Answer is:
C
Max. acceleration of `1kg`
`= ((0.6)(1)(10))/(1) = 6ms^(2)`
Max. acceleration of 2kg
`= ((0.4)(3)(10))/(3) = 4ms^(-2)`
Therefore maximum acceleration of system can be `4 m//s^(2)`
`rArr omega^(2)A =4 rArr A = (4)/(omega^(2)) = (4)/((k//m)) = (4)/(54//6) = 4/9m`
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Knowledge Check

  • Passage IV) Angular frequency in SHM is given by omega=sqrt(k/m) . Maximum acceleration in SHM is omega^(2) A and maximum value of friction between two bodies in contact is muN , where N is the normal reaction between the bodies. Now the value of k, the force constant is increased, then the maximum amplitude calcualted in above question will

    A
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    B
    increase
    C
    decrease
    D
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