56. If the pressure angle of the rack is 20°, the force acting along the line of action between the rack and the gear teeth is
    

1. 1. 250 N
      

   
   
   

   2. 342 N
      

   
   
   

   3. 532 N
      

   
   
   

   4. 600 N

   
   
   

---

1. View Hint View Answer Discuss in Forum

   P cos φ = F
   ∴ Force acting along the line of action,
   

   P =	F	=	500	= 532 N
   	cos φ		cos 20°

   Correct Option: C

   P cos φ = F
   ∴ Force acting along the line of action,
   

   P =	F	=	500	= 532 N
   	cos φ		cos 20°

---

57. A mass of 1 kg is suspended by means of 3 springs as shown in figure. The spring constant K₁, K₂ and K₃ are respectively 1 kN/m, 3 kN/m and 2 kN/m. The natural frequency of the system is approximately
    
    

1. 1. 46.90 rad/s
      

   
   
   

   2. 52.44 rad/s
      

   
   
   

   3. 60.55 rad/s
      

   
   
   

   4. 77.46 rad/s

   
   
   

---

1. View Hint View Answer Discuss in Forum

   K₁ and K₂ are in series,
   

   ∴	1	=	1	+	1	=	1	+	1	=	4
   	Keq1		K1		K2		1		3		3

   
   K_eq1 and K₃ are in parallel arrangement,
   

   Keq = Keq1 + K3 =	3	+ 2 =	11	kN / m
   	4		4

   
   Natural frequency, ω = √K_eq / m = √(11 × 10³) / (4 × 1) = 52.44 rad / s

   Correct Option: B

   K₁ and K₂ are in series,
   

   ∴	1	=	1	+	1	=	1	+	1	=	4
   	Keq1		K1		K2		1		3		3

   
   K_eq1 and K₃ are in parallel arrangement,
   

   Keq = Keq1 + K3 =	3	+ 2 =	11	kN / m
   	4		4

   
   Natural frequency, ω = √K_eq / m = √(11 × 10³) / (4 × 1) = 52.44 rad / s

---

---

58. Consider the system of two wagons shown in figure. The natural frequencies of this system are
    
    

1. 1. 0 , √2k / m
      

   
   
   

   2. √k / m , √2k / m
      

   
   
   

   3. √k / m , √k / 2m
      

   
   
   

   4. 0 , √k / 2m

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Natural frequencies will be ( 0 , √2k / m )
   
   For small displacement
   

   θ =	a	=	b
   	300		150

   
   Taking moment about hinged point ‘O’
   k.a (0.3) = Wb
   ⇒ K(0.3)² = 300 × 0.150
   

   K =	300 × 0.150	= 500 N / m
   	0.32

   
   

   Correct Option: A

   Natural frequencies will be ( 0 , √2k / m )
   
   For small displacement
   

   θ =	a	=	b
   	300		150

   
   Taking moment about hinged point ‘O’
   k.a (0.3) = Wb
   ⇒ K(0.3)² = 300 × 0.150
   

   K =	300 × 0.150	= 500 N / m
   	0.32

   
   

---

59. 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

   
   
   

---

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

   
   

---

---

60. 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

   
   
   

---

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

   
   

---