121. Considering massless rigid rod small oscillations, the natural frequency (in rad/s) of vibration of the system shown in the figure is
     
     
     

1. 1. √	400
      	1

      
      

   
   
   

   2. √	400
      	2

      
      

   
   
   

   3. √	400
      	3

      
      

   
   
   

   4. √	400
      	4

   
   
   

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1. View Hint View Answer Discuss in Forum

   Form ⇒ dx = 2rθ
   

   	d2x	= 2r θ̇̇
   	dt2

   
   Taking moments 2r θ.m. 2r.θ̇̇ + 400 × (rθ) × rθ = 0
   ⇒ 4m.θ̇̇ + 400 × θ = 0
   

   Correct Option: D

   Form ⇒ dx = 2rθ
   

   	d2x	= 2r θ̇̇
   	dt2

   
   Taking moments 2r θ.m. 2r.θ̇̇ + 400 × (rθ) × rθ = 0
   ⇒ 4m.θ̇̇ + 400 × θ = 0
   

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122. What is the natural frequency of the spring mass system shown below? The contact between the block and the inclined plane is frictionless. The mass of the block is denoted by m and the spring constants are denoted by k₁ and k₂ as shown below
     
     
     

1. 1. 
      

   
   
   

   2. 
      

   
   
   

   3. 
      

   
   
   

   4. 

   
   
   

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1. View Hint View Answer Discuss in Forum

   It is parallel
   k_eq = k₁ + k₂
   

   Correct Option: D

   It is parallel
   k_eq = k₁ + k₂
   

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123. A rigid uniform rod AB of length L and mass m is hinged at C such that AC = L/3, CB = 2L/3. Ends A and B are supported by springs of spring constant k. The natural frequency of the system is given by
     
     
     

1. 1. 
      

   
   
   

   2. 
      

   
   
   

   3. 
      

   
   
   

   4. 

   
   
   

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1. View Hint View Answer Discuss in Forum

   Kθ		L		2	+ Kθ		2L		2	+	mL2	θ̇̇ = 0
   		3					3				9

   
   

   ⇒		mL2		θ̇̇ +		5kL2		θ = 0
   		9				9

   
   

   θ̇̇ +		5k		θ = 0
   		m

   
   

   ωn =

   Correct Option: D

   Kθ		L		2	+ Kθ		2L		2	+	mL2	θ̇̇ = 0
   		3					3				9

   
   

   ⇒		mL2		θ̇̇ +		5kL2		θ = 0
   		9				9

   
   

   θ̇̇ +		5k		θ = 0
   		m

   
   

   ωn =

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124. A point mass is executing simple harmonic motion with an amplitude of 10 mm and frequency of 4 Hz. The maximum acceleration (m/s²) of the mass is _______

1. 1. 6.32 m/s²

   
   
   

   2. 61.32 m/s²

   
   
   

   3. 1.32 m/s²

   
   
   

   4. 10.32 m/s²

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given A = 10 mm = 0.01 m
   f = 4 Hz
   x = A cos ωt
   

   v =	dx	= -Aω sin ωt
   	dt

   
   

   	d2x	= a = -Aω2 cos ωt
   	dt2

   
   For maximum, t = 0
   ∴ a = Aω² = 0.01(2πf)²
   ∵ ω = 2πf = 0.01(2 × π × 4)² = 6.32 m/s²

   Correct Option: A

   Given A = 10 mm = 0.01 m
   f = 4 Hz
   x = A cos ωt
   

   v =	dx	= -Aω sin ωt
   	dt

   
   

   	d2x	= a = -Aω2 cos ωt
   	dt2

   
   For maximum, t = 0
   ∴ a = Aω² = 0.01(2πf)²
   ∵ ω = 2πf = 0.01(2 × π × 4)² = 6.32 m/s²

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125. A spur gear has pitch circle diameter D and number of teeth T. The circular pitch of the gear is
     

1. 1. D

      T

      
      

   
   
   

   2. πD

      T

      
      

   
   
   

   3. 2πD

      T

      
      

   
   
   

   4. T

      D

   
   
   

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1. View Hint View Answer Discuss in Forum

   Circular pitch , Pc =	πD
   	T

   Correct Option: B

   Circular pitch , Pc =	πD
   	T

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