VIBRATORY MOTION

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• Q1: Define Vibratory Motion.
  Ans: Vibratory motion is a to and fro motion of a body about its mean position.
• Q2: Define Mean Position.
  Ans: Mean position is the central position of a vibrating body where net force acting on it is zero.
• Q3: Define Simple Harmonic Motion.
  Ans: Simple Harmonic Motion (SHM) is a type of vibratory motion in which the restoring force is directly proportional to displacement and acts towards the mean position.
• Q4: State the characteristics of Simple Harmonic Motion.
  Ans: The characteristics of SHM are:

  • ➔ Motion is periodic in nature
  • ➔ Restoring force is proportional to displacement
  • ➔ Motion is about a fixed mean position
  • ➔ Acceleration is directed towards mean position
• Q5: Define Periodic Motion.
  Ans: Periodic motion is a motion that repeats itself after equal intervals of time.
• Q6: Define Displacement.
  Ans: Displacement is the distance of a particle from its mean position in a particular direction.
• Q7: Define Frequency.
  Ans: Frequency is the number of complete cycles of motion per second.
• Q8: Define Amplitude.
  Ans: Amplitude is the maximum displacement of a vibrating body from its mean position.
• Q9: Define Time Period.
  Ans: Time period is the time taken to complete one full cycle of motion.
• Q10: Define Angular Velocity.
  Ans: Angular velocity is the rate of change of angular displacement with respect to time.
• Q11: Define Velocity.
  Ans: Velocity is the rate of change of displacement with respect to time.
• Q12: What is a simple pendulum?
  Ans: A simple pendulum is a small heavy bob suspended by a light inextensible string from a fixed support.
• Q13: What is the time period of a simple pendulum?
  Ans: The time period of a simple pendulum is given by:
  T = 2π √(L/g)
  where L = length of pendulum, g = acceleration due to gravity
• Q14: What is the frequency of a simple pendulum?
  Ans: Frequency of a simple pendulum is:
  f = 1/T
  where T is the time period
• Q15: Define Seconds Pendulum.
  Ans: A seconds pendulum is a simple pendulum whose time period is 2 seconds.
• Q16: Define Conical Pendulum.
  Ans: A conical pendulum is a pendulum in which the bob moves in a horizontal circle forming a cone shape with the string.
• Q17: What is a Torsional pendulum?
  Ans: A torsional pendulum is a system where a disc or body is suspended by a wire and oscillates by twisting about its axis.
• Q18: Define Speed Governors.
  Ans: Speed governors are devices used to regulate the speed of an engine by controlling fuel supply.
• Q19: State the functions of Speed Governors.
  Ans: The functions of speed governors are:

  • ➔ Maintain constant engine speed
  • ➔ Control fuel supply according to load
  • ➔ Prevent over-speeding of engine
• Q20: State the types of Speed Governors.
  Ans: The types of speed governors are:

  • ➔ Centrifugal governor
  • ➔ Inertia governor
• Q21: Define Centrifugal Governor.
  Ans: A centrifugal governor is a device that uses centrifugal force to regulate engine speed.
• Q22: Define Inertia Governor.
  Ans: An inertia governor regulates engine speed based on the inertia forces acting due to speed changes.
• Q23: What is a Helical Spring? What is it used for?
  Ans: A helical spring is a coiled wire spring used to store energy and absorb shock.
  It is used in:

  • ➔ Vehicle suspension systems
  • ➔ Mechanical devices
  • ➔ Load absorption systems
• Q24: Define Cam.
  Ans: A cam is a mechanical device used to convert rotary motion into reciprocating or oscillating motion.
• Q25: State the types of Cam.
  Ans: The types of cam are:

  • ➔ Disc cam
  • ➔ Cylindrical cam
  • ➔ Wedge cam
• Q26: Define Disc Cam.
  Ans: A disc cam is a flat circular cam in which the follower moves due to its rotation.
• Q27: Define Cylindrical Cam.
  Ans: A cylindrical cam is a cam in which the follower moves along a groove cut on a cylindrical surface.
• Q28: What is Quick Return Motion Mechanism?
  Ans: A quick return motion mechanism is a mechanism in which the cutting tool takes less time in the return stroke than in the forward stroke.
• Q29: State the types of Quick Return Motion Mechanism.
  Ans: The types are:

  • ➔ Crank and slotted lever mechanism
  • ➔ Whitworth mechanism
• Q30: State the uses of Quick Return Motion Mechanism.
  Ans: The uses are:

  • ➔ Shaping machines
  • ➔ Slotting machines
  • ➔ Planers
• Q31: Calculate the maximum velocity and acceleration of the body if the amplitude of S.H.M is 10cm and frequency is 20 cycles/s.
  Ans: Given:
  Amplitude, A = 10 cm = 0.10 m
  Frequency, f = 20 cycles/s
  
  Step 1: Angular velocity
  Formula: ω = 2πf
  ω = 2π × 20
  ω = 40π rad/s
  
  Step 2: Maximum velocity
  Formula: V_max = ωA
  V_max = 40π × 0.10
  V_max = 4π m/s
  V_max = 4 × 3.14
  V_max = 12.56 m/s
  
  Step 3: Maximum acceleration
  Formula: a_max = ω²A
  a_max = (40π)² × 0.10
  a_max = 1600π² × 0.10
  a_max = 160π²
  a_max = 160 × 9.87
  a_max = 1579.2 m/s²
  
  Final Answers:
  Maximum velocity = 12.56 m/s
  Maximum acceleration = 1579.2 m/s²
• Q32: A simple pendulum is completing 2 vibrations per second. If g = 9.8 m/s², find the length of the pendulum.
  Ans: Given:
  Frequency, f = 2 vibrations/s
  Acceleration due to gravity, g = 9.8 m/s²
  
  Step 1: Find Time Period
  Formula: T = 1/f
  T = 1/2
  T = 0.5 s
  
  Step 2: Formula of simple pendulum
  T = 2π √(L/g)
  
  Step 3: Rearranging formula for L
  T² = 4π²(L/g)
  L = (g T²) / (4π²)
  
  Step 4: Substituting values
  L = (9.8 × (0.5)²) / (4π²)
  L = (9.8 × 0.25) / (4π²)
  L = 2.45 / (4π²)
  
  Step 5: Calculation
  π² ≈ 9.87
  4π² ≈ 39.48
  
  L = 2.45 / 39.48
  L ≈ 0.062 m
  
  Final Answer:
  Length of pendulum ≈ 0.062 m = 6.2 cm
• Q33: The buffer of a railway wagon is compressed 5 cm when a force of 200 N is applied to it. Calculate the spring constant.
  Ans: Given:
  Force, F = 200 N
  Compression, x = 5 cm = 0.05 m
  
  Step 1: Formula of Spring Constant
  F = kx
  k = F / x
  
  Step 2: Substituting values
  k = 200 / 0.05
  
  Step 3: Calculation
  k = 4000 N/m
  
  Final Answer:
  Spring constant, k = 4000 N/m
• Q34: A body of mass 0.5 kg is attached to a spring having spring constant of 8 N/m. Find its time period.
  Ans: Given:
  Mass, m = 0.5 kg
  Spring constant, k = 8 N/m
  
  Step 1: Formula of time period of spring system
  T = 2π √(m/k)
  
  Step 2: Substituting values
  T = 2π √(0.5 / 8)
  T = 2π √(0.0625)
  
  Step 3: Simplify square root
  √0.0625 = 0.25
  
  T = 2π × 0.25
  T = 0.5π s
  
  Step 4: Approximation
  π ≈ 3.14
  T = 0.5 × 3.14
  T = 1.57 s
  
  Final Answer:
  Time period, T = 1.57 s