11. 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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1. View Hint View Answer Discuss in Forum

   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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12. 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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1. View Hint View Answer Discuss in Forum

   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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13. Two simple harmonic motions with the same frequency act on a particle at right angles i.e., along x and y axis. If the two amplitudes are equal and the phase difference is π/2, the resultant motion will be​​
    

1. 1. a circle

   
   
   

   2. an ellipse with the major axis along y-axis

   
   
   

   3. an ellipse with the major axis along x-axis

   
   
   

   4. a straight line inclined at 45° to the x-axis

   
   
   

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1. View Hint View Answer Discuss in Forum

   Equation of two simple harmonic motions
   y = A sin (ωt + φ) .....................(1)
   

   x = A sin		ωt + φ +	π
   			2

   
   ⇒ x = A cos (ωt + φ) .....................(2)
   On squaring and adding equations  (1) and (2), ​x² + y² = A² ​
   This is an equation of a circle. Hence, resulting motion will be a circular motion.
   

   Correct Option: A

   Equation of two simple harmonic motions
   y = A sin (ωt + φ) .....................(1)
   

   x = A sin		ωt + φ +	π
   			2

   
   ⇒ x = A cos (ωt + φ) .....................(2)
   On squaring and adding equations  (1) and (2), ​x² + y² = A² ​
   This is an equation of a circle. Hence, resulting motion will be a circular motion.
   

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14. 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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1. View Hint View Answer Discuss in Forum

   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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15. The potential energy of a simple harmonic oscillator when the particle is half way to its end point is

1. 1. 	1	E
      	2

   
   
   

   2. 	2	E
      	3

   
   
   

   3. 	1	E
      	8

   
   
   

   4. 	1	E
      	4

      
      where E is the total energy

   
   
   

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1. View Hint View Answer Discuss in Forum

   P.E. =	1	kx2 = E
   	2

   
   At half way
   

   P.E. =	1	k		x		2	=	1	kx2		=	E
   	2			2				2				4
   								4

   
   

   Correct Option: D

   P.E. =	1	kx2 = E
   	2

   
   At half way
   

   P.E. =	1	k		x		2	=	1	kx2		=	E
   	2			2				2				4
   								4

   
   

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