JEE Physics/ Lakshmikanta Satapathy/ Simple harmonic motion QA part 6/ Question on amplitude of SHM determined from its total energy solved with the related concepts

1. Physics Helpline
L K Satapathy
Oscillations 6
Energy in SHM & Amplitude
2 21
2
E K U m A  

2. Physics Helpline
L K Satapathy
Oscillations 6
Concepts :
Question : A particle of mass 200 grams executes Simple Harmonic Motion along
the x-axis with time period  /25 seconds . At x = 4 cm , its kinetic energy is 0.5 J
and potential energy is 0.4 J. Then its amplitude is
(a) 4 cm (b) 5 cm (c) 6 cm (d) 25 cm
For a Particle of mass m , executing S H M of frequency f , angular frequency  ,
time period = T and amplitude = A
1 2
2f f
T T

    
At displacement x , Kinetic energy 2 2 21
( )
2
K m A x 
Potential energy 2 21
2
U m x
 Total energy 2 21
2
E K U m A  

3. Physics Helpline
L K Satapathy
Oscillations 6
Correct option = (c)
2
50
25
T rad s
T
 
   
Answer :
2
2
2 2 0.9 9
0.2 2500 2500
E
A
m

   

3
6
50
A m cm  
Given :
 E = K + U = 0 . 9 J
0.2 , 0.5 , 0.4 , sec
25
m kg K J U J T

   
2 21
2
E m A

4. Physics Helpline
L K Satapathy
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