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 MotionPeriodic 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 = -2xor

Measures of SHM

Amplitude (A): Maximum displacement from equilibrium position.

0x x Ax A

Phase: The argument of sine function in equation of SHM is called phase.

SHM and Circular Motion

t

x = A sint

v = A cost

a = -A 2 sint

a = -2x

Asint

AcostA

2 2Hence, v A x

As x = A sint,

Measures of SHMSince,

hence, ma = -kx.

a = -2x

F = -kx

xmka Also,

k angular frequencym

1 kff requency2 2 m

2 mT 2 Time periodk

Energy Variation in SHM

22Am21PE

0KE

x A 0x

0PEAm2

1KE 22

22Am2

1PE

0KE

x A

PE

KE

2 2 2

2 2

2 2

At any x1KE m (A x )21PE m x2

1TE PE KE m A2

Class Test

Class Exercise - 1A 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

T 2T(a) (b)12 12

3T 6T(c) (d)12 12

Solution Let x = A sin(t + )

Here as at t = 0 particle is at extreme

2

2x A sin t

T 2

A

But x2

A 2A sin t

2 T 2

1 2or sin t

2 T 2

1 5But sin

2 6

5 2So t

6 T 2

2T

or t12

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

T T(a) , , (b) T, T, 2 2

T T T(c) , , (d) None of these2 2 2

Solution 2 21

PE m X2

2 2 21m A sin ( t )

2

2 21 1– cos2( t )

m A2 2

2 21 2m A 1– cos 2 t

4 T

So time period of PE is . Similarly, proceeding we get the result for KE.

T2

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 forceF = –(abc)2x, what will be it’s time period?

2

M 2(a) 2 (b) Mabc abc

2M(c) 2 (d) None of these(abc)

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

2

MT 2

(abc)

Hence answer is (b).

Class Exercise - 4Amplitude of particle whose equation of motion is represented as is

(a) 5 (b) 4(c) 3 (d) Cannot solve

SolutionLet A be the amplitude and the initial phase, then A[sin314t cos cos314t sin ]

A cos sin314t A sin cos314t

But x 3sin314t 4cos314t

A cos 3 .............(i)

A sin 4 ..................(ii)

Squaring and adding, we getA2 = 25 or A = 5 Hence answer is (a).

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

1 1

1 1

4 3(a) tan (b) tan3 4

4 5(c) tan (d) tan5 4

SolutionDividing equation (ii) by equation (i), we get

4

tan3

1 4tan

3

Hence answer is (a).

Class Exercise - 6

Phase difference between acceleration and velocity of SHM is . (True/False)

SolutionIf equation of SHM is x A sin t

Then, v A sin t2

2and acceleration a A sin( t )

So phase difference

( t ) – t2

2

So false.

Class Exercise - 7Keeping 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

2 2(i)

1initial total energy E m A

2

2 2(i)

1or E mA (2 f )

2

2 2 21mA 4 f

2

2 2 2f

1E mA 4 (2f )

2

i4 E

f iSo E – E 3EHence answer is (a).

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

A2

E 3E(a) (d)

4 4

E(c) (d) zero2

Solution

2 2(x)

1PE m x

2

2 21 ATotal energy E m A at x

2 2

22

(A/2)1 A

PE m2 2

2

21 Am

2 4

E4

Hence answer is (a).

Class Exercise - 9If E is the total energy and A is the amplitude of SHM, then

A 2

E E

A

A

E

A

E

(a) (b)

(c) (d)

Solution

2 21Total energy (E) m A2

E A2

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)?

A A(a) (b) –2 2

A A(c) (d) –2 2

Solution

2 21PE m x

2

2 2 21KE m (A – x )

2

2 2 2 2 21 1m X m (A – x )

2 2

PE = KE

2 2 2or x A – x

2 2or A 2x

Aor x2

where A is amplitude of SHM.Hence answer is (a, b).

Thank you