physics. simple harmonic motion - 1 session session objectives
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Session ObjectivesTRANSCRIPT
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