Physics

Simple Harmonic Motion - 1 Session

Session Objectives

Session Objective SHM - Concept & Cause Relationship between parameters & motion Displ. Vs Time relationship for SHM SHM as a projection of circular motion Energy of a SHM Oscillator

Periodic and Oscillatory Motion Periodic or harmonic motion: motion which repeats itself after regular interval. Oscillatory or vibratory motion: periodic motion on same path (to and fro motion) between fixed limits.

Simple Harmonic Motion  Restoring force is always directed towards the mean position. F = -kx  Oscillatory motion within fixed limits. a = - 2 x or

Measures of SHM Amplitude (A): Maximum displacement from equilibrium position. Phase: The argument of sine function in equation of SHM is called phase.

SHM and Circular Motion tt x = A sint v = A  cost a = -A  2 sint a = - 2 x Asint Acost A As x = A sint,

Measures of SHM Since, hence, ma = -kx. a = - 2 x F = -kx Also,

Energy Variation in SHM PE KE

Class Test

Class Exercise - 1 A particle is executing SHM of amplitude A with time period T in second. The time taken by it to move from positive extreme position to half the amplitude is

Solution Let x = A sin(  t +  ) Here as at t = 0 particle is at extreme

Class Exercise - 2 A particle executes an undamped SHM of time period T. Then the time period with which PE, KE and TE changes is respectively

Solution So time period of PE is. Similarly, proceeding we get the result for KE. As total energy is constant so it oscillates with time period infinity. Hence answer is (a).

Class Exercise - 3 A body of mass M experience a force F = –(abc) 2 x, what will be it’s time period?

Solution As force is directly proportional to X and directed towards mean position (due to negative sign of force), the particle will execute SHM. The force constant of SHM would be K = (abc) 2 So time period Hence answer is (b).

Class Exercise - 4 Amplitude of particle whose equation of motion is represented as is (a) 5(b) 4 (c) 3(d) Cannot solve

Solution Let A be the amplitude and  the initial phase, then Squaring and adding, we get A 2 = 25 or A = 5 Hence answer is (a).

Class Exercise - 5 What will be the initial phase () in problem 4?

Solution Dividing equation (ii) by equation (i), we get Hence answer is (a).

Class Exercise - 6 Phase difference between acceleration and velocity of SHM is . (True/False)

Solution If equation of SHM is So phase difference So false.

Class Exercise - 7 Keeping amplitude same if frequency of the source is changed from f to 2f. Then total energy is changed by (a) 3E(b) E (c) zero(d) 4E

Solution Hence answer is (a).

Class Exercise - 8 If E is the total energy of SHM, what is the value of PE at where A is the amplitude of SHM?

Solution Hence answer is (a).

Class Exercise - 9 If E is the total energy and A is the amplitude of SHM, then (a) (b) (c) (d)

Solution E  A 2 Hence (a) and (b) is appropriate graph. Hence answer is (a, b).

Class Exercise - 10 At what displacement x, PE is equal to KE (for SHM)?

Solution PE = KE where A is amplitude of SHM. Hence answer is (a, b).

Thank you