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

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?

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?

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?

Statement - 1 : Maximum value of friction force between two surfaces is mu xx" normal reaction" . where mu = coefficient of friction between surfaces. Statement - 2 : Friction force between surfaces of two bodies is always less than or equal to externally applied force.

If the angular frequency of a particle executing SHM is omega and its maximum velocity is v_(m) , then show that the relation between its velocity v and acceleration a is a=-omegasqrt(v_(m)^(2)-v^(2)) .

For a body in S.H.M the velocity is given by the relation v = sqrt(144-16 x^(2))m//sec . The maximum acceleration is