136. For an under damped harmonic oscillator, resonance
     

1. 1. occurs when excitation frequency is greater than undamped natural frequency
      

   
   
   

   2. occurs when excitation frequency is less than undamped natural frequency
      

   
   
   

   3. occurs when excitation frequency is equal to undamped natural frequency
      

   
   
   

   4. never occurs

   
   
   

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1. View Hint View Answer Discuss in Forum

   For underdamped harmonic oscillator,
   ω_d = ω_n

   Correct Option: C

   For underdamped harmonic oscillator,
   ω_d = ω_n

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137. A uniform stiff rod of length 300 mm and having a weight of 300 N is pivoted at one end and connected to a spring at the other end. For keeping the rod vertical in a stable position the minimum value of spring constant K needed is
     
     

1. 1. 300 N/m
      

   
   
   

   2. 400 N/m
      

   
   
   

   3. 500 N/m
      

   
   
   

   4. 1000 N/m

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: C

   NA

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138. The assembly shown in the figure is composed of two massless rods of length l with two particles, each of mass m. The natural frequency of this assembly for small oscillations is
     
     

1. 1. √g / l
      

   
   
   

   2. √2g / l(cosα)
      

   
   
   

   3. √g / l(cosα)
      

   
   
   

   4. √(gcosα) / l

   
   
   

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1. View Hint View Answer Discuss in Forum

   Net restoring torque when displaced by a small angle θ
   τ = mg cos (α – θ) l – mg (α + θ)l
   = 2 mgl cos α sin θ
   For very small θ, sin θ ≈ θ
   ∴ τ = 2mgl cos α . θ (restorative)
   

   Now , I	d2θ	+ 2mgl cosα . θ = 0
   	dt2

   
   But I = 2ml²
   

   ∴ 2ml2	d2θ	+ 2mgl cosα . θ = 0
   	dt2

   
   

   or	d2θ	+	g cosα	θ = 0
   	dt2		l

   
   

   ∴ ωn = √	g cosα
   	l

   
   

   Correct Option: D

   Net restoring torque when displaced by a small angle θ
   τ = mg cos (α – θ) l – mg (α + θ)l
   = 2 mgl cos α sin θ
   For very small θ, sin θ ≈ θ
   ∴ τ = 2mgl cos α . θ (restorative)
   

   Now , I	d2θ	+ 2mgl cosα . θ = 0
   	dt2

   
   But I = 2ml²
   

   ∴ 2ml2	d2θ	+ 2mgl cosα . θ = 0
   	dt2

   
   

   or	d2θ	+	g cosα	θ = 0
   	dt2		l

   
   

   ∴ ωn = √	g cosα
   	l

   
   

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139. In the figure shown, the spring deflects by δ to position A (the equilibrium position) when a mass m is kept on it. During free vibration, the mass is at position B at some instant. The change in potential energy of the spring mass system from position A to position B is
     
     

1. 1. 1	kx2
      2

      
      

   
   
   

   2. 1	kx2 - mgx
      2

      
      

   
   
   

   3. 1	k(x + δ)
      2

      
      

   
   
   

   4. 1	kx2 + mgx
      2

   
   
   

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1. View Hint View Answer Discuss in Forum

   Potential energy at A = mg (l – δ)
   

   Total energy at B = mg [l – (δ + x)] +	1	kx2
   	2

   
   

   ∴ Change in energy = mgl – mg(δ + x) +	1	kx2 - mgl + mgδ
   	2

   
   

   =	1	kx2 - mgx . δ
   	2

   
   

   Correct Option: A

   Potential energy at A = mg (l – δ)
   

   Total energy at B = mg [l – (δ + x)] +	1	kx2
   	2

   
   

   ∴ Change in energy = mgl – mg(δ + x) +	1	kx2 - mgl + mgδ
   	2

   
   

   =	1	kx2 - mgx . δ
   	2

   
   

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140. As shown in figure, a mass of 100 kg is held between two springs. The natural frequency of vibration of the system in cycle/s, is
     
     

1. 1. 1

      2π

      
      

   
   
   

   2. 5

      π

      
      

   
   
   

   3. 10

      π

      
      

   
   
   

   4. 20

      π

   
   
   

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1. View Hint View Answer Discuss in Forum

   S = S₁ + S₂ = 20 + 20 = 40 kN/m = 40,000 N/m
   ∴ Natural frequency of vibration of the system,
   

   fn =	1	√S / m
   	2π

   
   

   fn =	1	√(400 × 1000)/ 100	=	20
   	2π			2π

   
   

   fn =	10
   	π

   
   

   Correct Option: C

   S = S₁ + S₂ = 20 + 20 = 40 kN/m = 40,000 N/m
   ∴ Natural frequency of vibration of the system,
   

   fn =	1	√S / m
   	2π

   
   

   fn =	1	√(400 × 1000)/ 100	=	20
   	2π			2π

   
   

   fn =	10
   	π

   
   

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