21. A particle is executing a simple harmonic motion of amplitude a. Its potential energy is maximum when the displacement from the position of the maximum kinetic energy is
    

1. 1. 0

   
   
   

   2. -a

   
   
   

   3. a / 2

   
   
   

   4. – a / 2

   
   
   

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   P.E. of particle executing S.H.M. =	1	mω2 x2
   	2

   
   

   At x = a, P.E. is maximum i.e. =	1	mω2 a2
   	2

   
   At x = 0, K.E. is maximum. Hence, displacement from position of maximum Kinetic energy = a.

   Correct Option: B

   P.E. of particle executing S.H.M. =	1	mω2 x2
   	2

   
   

   At x = a, P.E. is maximum i.e. =	1	mω2 a2
   	2

   
   At x = 0, K.E. is maximum. Hence, displacement from position of maximum Kinetic energy = a.

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22. There is a body having mass m and performing S.H.M. with amplitude a. There is a restoring force F = –kx. The total energy of body depends upon​​​​
    

1. 1. k, x

   
   
   

   2. k, a

   
   
   

   3. ​k, a, x

   
   
   

   4. k, a, v

   
   
   

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   Total Energy of body performing simple harmonic motion =	1	mω2 a2 =	1	ka2
   	2		2

   
   where k = mω²
   

   	T.E. = K.E. + P.E. =	1	m(a2 - x2) ω2 +	1	mω2 x2
   		2		2

   
   Hence energy depends upon amplitude and k (spring constant).

   Correct Option: B

   Total Energy of body performing simple harmonic motion =	1	mω2 a2 =	1	ka2
   	2		2

   
   where k = mω²
   

   	T.E. = K.E. + P.E. =	1	m(a2 - x2) ω2 +	1	mω2 x2
   		2		2

   
   Hence energy depends upon amplitude and k (spring constant).

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23. In a simple harmonic motion, when the displacement is one-half the amplitude, what fraction of the total energy is kinetic?​
    

1. 1. 0

   
   
   

   2. 	1
      	4

   
   
   

   3. 	1
      	2

   
   
   

   4. 	3
      	4

   
   
   

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   Total energy of particle executing S.H.M. of amplitude (A).
   

   E =	1	mω2 A2
   	2

   
   

   K.E.of the particle =	1	mω2		A2 -	A2			when x =	A
   	2				4				2

   
   

   =	1	mω2 ×	3	A2 =	1	×	3	mω2 A2
   	2		4		2		4

   
   

   Clearly ,	KE	=	3
   	Total energy		4

   Correct Option: D

   Total energy of particle executing S.H.M. of amplitude (A).
   

   E =	1	mω2 A2
   	2

   
   

   K.E.of the particle =	1	mω2		A2 -	A2			when x =	A
   	2				4				2

   
   

   =	1	mω2 ×	3	A2 =	1	×	3	mω2 A2
   	2		4		2		4

   
   

   Clearly ,	KE	=	3
   	Total energy		4

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24. A linear harmonic oscillator of force constant 2 × 10⁶ N/m and amplitude 0.01 m has a total mechanical energy of 160 J. Its​
    

1. 1. potential energy is 160 J

   
   
   

   2. potential energy is 100 J

   
   
   

   3. potential energy is zero

   
   
   

   4. potential energy is 120 J

   
   
   

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   Force constant k = 2 × 10⁶ N/m ​
   Amplitude (x) = 0.01 m ​
   

   Potential Energy =	1	kx2
   	2

   
   

   =	1	× (2 × 106) × (0.01)2 = 100 J
   	2

   Correct Option: B

   Force constant k = 2 × 10⁶ N/m ​
   Amplitude (x) = 0.01 m ​
   

   Potential Energy =	1	kx2
   	2

   
   

   =	1	× (2 × 106) × (0.01)2 = 100 J
   	2

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25. A body executes S.H.M with an amplitude A. At what displacement from the mean position is the potential energy of the body is one fourth of its total energy ?​​​
    

1. 1. A / 4

   
   
   

   2. A / 2 ​

   
   
   

   3. 3A / 4

   
   
   

   4. Some other fraction of A.

   
   
   

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1. View Hint View Answer Discuss in Forum

   P.E. , V =	1	mω2 x2
   	2

   
   

   Total energy , E =	1	mω2 A2
   	2

   
   

   P.E. =	1	E ⇒	1	mω2 x2 =	1	mω2 A2
   	4		2		8

   
   

   ∴ x =	1	A
   	2

   
   

   Correct Option: B

   P.E. , V =	1	mω2 x2
   	2

   
   

   Total energy , E =	1	mω2 A2
   	2

   
   

   P.E. =	1	E ⇒	1	mω2 x2 =	1	mω2 A2
   	4		2		8

   
   

   ∴ x =	1	A
   	2

   
   

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