Home
Class 12
PHYSICS
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.
Now the value of k, the force constant is increased, then the maximum amplitude calcualted in above question will

A

remain same

B

increase

C

decrease

D

data in insufficient

Text Solution

Verified by Experts

The correct Answer is:
C
`omega^(2) = "constant" rArr A prop 1/k`
Promotional Banner

Topper's Solved these Questions

  • SIMPLE HARMONIC MOTION

    ALLEN|Exercise Comprehension Base Questions (2)|2 Videos

  • SIMPLE HARMONIC MOTION

    ALLEN|Exercise Comprehension Base Questions (3)|2 Videos

  • SIMPLE HARMONIC MOTION

    ALLEN|Exercise Asseration & Reason|7 Videos

  • RACE

    ALLEN|Exercise Basic Maths (Wave Motion & Dopplers Effect) (Stationary waves & doppler effect, beats)|25 Videos

  • TEST PAPER

    ALLEN|Exercise PHYSICS|4 Videos

Similar Questions

Explore conceptually related problems

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.

Two particles are in SHM with the same amplitude and frequency along the same line and about the same point. If the maximum separation between them is sqrt(3) times their amplitude, the phase difference between them is (2pi)/n . Find value of n

Prism 1 with bar 2 of mass m placed on it gets a horizontal acceleration w directed to the left (figure). At what maximum value of this acceleration will the bar still stationary relative to the prism, if the coefficient of friction between them k lt cot alpha ?

In the previous question, the angular frequency of the simple harmonic motion is omega . The coefficient of friction between the coin and the platform is mu . The amplitude of oscillation is gradually increased. The coin will begin to slip on the platform for the first time (i) at the extreme positions of oscillations (ii) at the mean position (iii) for an amplitude of (mug)/(omega^2) (iv) for an amplitude of (g)/(muomega^2)

A body is vibrating with SHM of amplitude 15 cm and frequency 4 Hz. Compute the maximum values of the acceleration and velocity.

The system shown in the figure is in equilibrium The maximum value of W, so that the maximum value of static frictional force on 100kg. body is 450N, will be:

The angular frequency of the damped oscillator is given by omega=sqrt((k)/(m)-(r^(2))/(4m^(2))) ,where k is the spring constant, m is the mass of the oscillator and r is the damping constant. If the ratio r^(2)//(m k) is 8% ,the change in the time period compared to the undamped oscillator