51. A mass of 2.0 kg is put on a flat pan attached to a vertical spring fixed on the ground as shown in the figure. The mass of the spring and the pan is negligible. When pressed slightly and released the mass executes a simple harmonic motion. The spring constant is 200 N/m. What should be the minimum amplitude of the motion so that the mass gets detached from the pan (take g = 10 m/s²)?​​
    
    

1. 1. 10.0 cm

   
   
   

   2. any value less than 12.0 cm

   
   
   

   3. 4.0 cm

   
   
   

   4. 8.0 cm

   
   
   

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1. View Hint View Answer Discuss in Forum

   Mass gets detached at the upper extreme position when pan returns to its mean position. ​
   At that point, R = mg – mω²a = 0 ​
   i.e. g = ω²a
   ⇒ a = g / ω² = mg / k
   

   ⇒ a =	2 × 10			As ω2 =	k
   	200				m

   
   ⇒ a = (1 / 10) m = 10 cm

   Correct Option: A

   Mass gets detached at the upper extreme position when pan returns to its mean position. ​
   At that point, R = mg – mω²a = 0 ​
   i.e. g = ω²a
   ⇒ a = g / ω² = mg / k
   

   ⇒ a =	2 × 10			As ω2 =	k
   	200				m

   
   ⇒ a = (1 / 10) m = 10 cm

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52. ​A point performs simple harmonic oscillation of period T and the equation of motion is given by x = a sin (ωt + π/6). After the elapse of what fraction of the time period the velocity of the point will be equal to half of its maximum velocity?
    

1. 1. T/8

   
   
   

   2. T/6

   
   
   

   3. T/3

   
   
   

   4. T/12

   
   
   

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1. View Hint View Answer Discuss in Forum

   We have , x = a sin		ωt +	π
   			6

   
   

   ∴ Velocity , v =	dx	= aω cos		ωt +	π
   	dt				6

   
   Maximum velocity = aω
   ​According to question,
   

   	aω	= aω cos		ωt +	π
   	2				6

   
   

   or cos		ωt +	π		=	1	= cos 60° or cos	π
   			6			2		3

   
   

   ⇒ ωt +	π	=	π
   	6		3

   
   

   ⇒ ωt =	π	-	π	or ωt =	π
   	3		6		6

   
   

   or	2π	.t =	π	⇒ t =	T
   	T		6		12

   Correct Option: D

   We have , x = a sin		ωt +	π
   			6

   
   

   ∴ Velocity , v =	dx	= aω cos		ωt +	π
   	dt				6

   
   Maximum velocity = aω
   ​According to question,
   

   	aω	= aω cos		ωt +	π
   	2				6

   
   

   or cos		ωt +	π		=	1	= cos 60° or cos	π
   			6			2		3

   
   

   ⇒ ωt +	π	=	π
   	6		3

   
   

   ⇒ ωt =	π	-	π	or ωt =	π
   	3		6		6

   
   

   or	2π	.t =	π	⇒ t =	T
   	T		6		12

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53. A block of mass M is attached to the lower end of a vertical spring. The spring is hung from a ceiling and has force constant value k. The mass is released from rest with the spring initially unstretched. The maximum extension produced in the length of the spring will be :​
    

1. 1. 2 Mg/k

   
   
   

   2. 4 Mg/k

   
   
   

   3. Mg/2k

   
   
   

   4. Mg/k

   
   
   

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1. View Hint View Answer Discuss in Forum

   Restoring force, f ′ = – kx ​
   where x is the extension produced in the spring. ​
   Weight of the mass acting downward = Mg. ​
   

   In equilibrium ​kx = Mg  or x =	Mg
   	k

   Correct Option: D

   Restoring force, f ′ = – kx ​
   where x is the extension produced in the spring. ​
   Weight of the mass acting downward = Mg. ​
   

   In equilibrium ​kx = Mg  or x =	Mg
   	k

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54. A simple pendulum performs simple harmonic motion about x = 0 with an amplitude a and time period T. The

    speed of the pendulum at x =	a	will be : ​​
    	2

1. 1. 	πa
      	T

   
   
   

   2. 	3π2 a
      	T

   
   
   

   3. 	πa √3
      	T

   
   
   

   4. 	πa √3
      	2T

   
   
   

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1. View Hint View Answer Discuss in Forum

   ​

   Speed v = ω √a² - x² , x =	a
   	2

   
   

   ∴ V = ω√{a² - (a² / 4)} = ω
   		3a2
   		4

   
   

   =	2π		a √3	=	πa √3
   	T		2		T

   
   

   Correct Option: C

   ​

   Speed v = ω √a² - x² , x =	a
   	2

   
   

   ∴ V = ω√{a² - (a² / 4)} = ω
   		3a2
   		4

   
   

   =	2π		a √3	=	πa √3
   	T		2		T

   
   

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55. The period of oscillation of a mass M suspended from a spring of negligible mass is T. If along with it another mass M is also suspended, the period of oscillation will now be​
    

1. 1. T

   
   
   

   2. T / √2

   
   
   

   3. 2T ​

   
   
   

   4. √2 T

   
   
   

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1. View Hint View Answer Discuss in Forum

   T = 2π
   		m
   		K

   
   

   ∴	T1	=
   				M1
   	T2			M2

   
   

   ∴ T2 = T1			= T1
   		M2			2M
   		M1			M

   
   T₂ = T₁ √2 = √2 T (where T₁ = T)

   Correct Option: D

   T = 2π
   		m
   		K

   
   

   ∴	T1	=
   				M1
   	T2			M2

   
   

   ∴ T2 = T1			= T1
   		M2			2M
   		M1			M

   
   T₂ = T₁ √2 = √2 T (where T₁ = T)

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