131. A disc of mass m is attached to a spring of stiffness k as shown in the fig. The disc rolls without slipping on a horizontal surface. The natural frequency of vibration of the system is
     
     

1. 1. 1	√k / m
      2π

      
      

   
   
   

   2. 1	√2k / m
      2π

      
      

   
   
   

   3. 1	√2k / 3m
      2π

      
      

   
   
   

   4. 1	√3k / 2m
      2π

      
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   Taking moments about instantaneous centre ‘A’, I_aθ̇̇ + (kx) r = 0
   ⇒ (I_o + mr²) θ̇̇ + kx (θr)r = 0
   

   ⇒		1	mr2 + mr2		θ	+ k(θr2) = 0
   		2

   
   

   ⇒ θ̇̇ +	kr2	θ = 0
   	3mr2
   	2

   
   

   ⇒ θ̇̇ +	2k	θ = 0
   	3m

   
   

   ∴ ωn =	1	√2k / 3m
   	2π

   
   

   Correct Option: C

   
   Taking moments about instantaneous centre ‘A’, I_aθ̇̇ + (kx) r = 0
   ⇒ (I_o + mr²) θ̇̇ + kx (θr)r = 0
   

   ⇒		1	mr2 + mr2		θ	+ k(θr2) = 0
   		2

   
   

   ⇒ θ̇̇ +	kr2	θ = 0
   	3mr2
   	2

   
   

   ⇒ θ̇̇ +	2k	θ = 0
   	3m

   
   

   ∴ ωn =	1	√2k / 3m
   	2π

   
   

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132. The natural frequency of a spring-mass system on earth is ω_n . The natural frequency of this system on the moon (g_moon = g_earth / 6) is
     

1. 1. ω_n
      

   
   
   

   2. 0.408 ω_n
      

   
   
   

   3. 0.204 ω_n
      

   
   
   

   4. 0.167 ω_n

   
   
   

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1. View Hint View Answer Discuss in Forum

   Natural frequency, ωn = √	k
   	2m

   
   I t depends only on mass but not on weight. Therefore it is independent of gravity of a system, so it will remain ω_n .

   Correct Option: A

   Natural frequency, ωn = √	k
   	2m

   
   I t depends only on mass but not on weight. Therefore it is independent of gravity of a system, so it will remain ω_n .

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133. A uniform rigid rod of mass m = 1 kg and length L = 1 m is hinged at its centre & laterally supported at one end by a spring of spring constant k = 300 N/m. The natural frequency ω_n in rad/s is
     

1. 1. 10
      

   
   
   

   2. 20
      

   
   
   

   3. 30
      

   
   
   

   4. 40

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   

   	1	Iθ̇ 2 +	1	kx2 = Constant
   	2		2

   
   

   	1	Iθ̇ 2 +	1	k		L	θ		2	= Constant
   	2		2			2

   
   Differentiating above equation w.r.t. t, we get
   

   	1	Iθ̇̇ +	kL2θ	= 0
   	2		4

   
   

   I =	mL2
   	12

   
   

   ⇒	mL2	θ̇̇ +	kL2	θ = 0
   	12		4

   
   

   θ̇̇ +	3k	θ = 0
   	m

   
   ω_n = √3k / m = √(3 × 300) / 1 = 30 rad/s
   

   Correct Option: C

   
   

   	1	Iθ̇ 2 +	1	kx2 = Constant
   	2		2

   
   

   	1	Iθ̇ 2 +	1	k		L	θ		2	= Constant
   	2		2			2

   
   Differentiating above equation w.r.t. t, we get
   

   	1	Iθ̇̇ +	kL2θ	= 0
   	2		4

   
   

   I =	mL2
   	12

   
   

   ⇒	mL2	θ̇̇ +	kL2	θ = 0
   	12		4

   
   

   θ̇̇ +	3k	θ = 0
   	m

   
   ω_n = √3k / m = √(3 × 300) / 1 = 30 rad/s
   

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134. The natural frequency of the spring mass system shown in the figure is closest to
     
     

1. 1. 8Hz
      

   
   
   

   2. 10Hz
      

   
   
   

   3. 12 Hz
      

   
   
   

   4. 14 Hz

   
   
   

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1. View Hint View Answer Discuss in Forum

   k_equ = k₁ + k₂ = 5600 N / m
   

   ∴ f =	1	√k / m = 10 Hz
   	2π

   
   

   Correct Option: B

   k_equ = k₁ + k₂ = 5600 N / m
   

   ∴ f =	1	√k / m = 10 Hz
   	2π

   
   

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135. The natural frequency of the system shown below
     
     

1. 1. √k / 2m
      

   
   
   

   2. √k / m
      

   
   
   

   3. √2k / m
      

   
   
   

   4. √3k / m

   
   
   

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1. View Hint View Answer Discuss in Forum

   Equivalent k,
   

   kequ =	1	+	1	=	k
   	k		k / 2 + k / 2		2

   
   

   ∴ ωn = √	k
   	2m

   Correct Option: A

   Equivalent k,
   

   kequ =	1	+	1	=	k
   	k		k / 2 + k / 2		2

   
   

   ∴ ωn = √	k
   	2m

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