Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
Concepts Of Physics MCQ Edition [Volume 1]PhysicsSimple Harmonic Motion

Assume all surfaces depicted in the diagram lack friction. The car has a mass M , the block has a mass m , and the spring possesses a force constant k . The system is released from rest with the spring initially stretched by a distance x₀ . Determine the time period of the resulting simple harmonic motions.

Options

  1. A2 M k
  2. B2 mM k(M + m)
  3. C2 M + m k
  4. D2 m k

Correct answer

B. 2 mM k(M + m)

Step-by-step solution

Let x₁ and x₂ be the displacements of the block of mass m and the car of mass M from their respective equilibrium positions. Since no external horizontal force acts on the system, the center of mass remains stationary. Therefore: m x₁ + M x₂ = 0 x₂ = - m M x₁ The net extension or compression in the spring is the relative displacement x = x₁ - x₂ . Substituting the value of x₂ , we get: x = x₁ - (- m M x₁ ) = x₁ (1 + m M ) = x₁ ( M + m M ) The restoring force acting on the block of mass m is F = -kx . F = -k ( M + m

Practice Simple Harmonic Motion on Quantrex Academy →

More from Simple Harmonic Motion

At a specific instant, the magnitudes of the position, velocity, and acceleration of a particle undergoing simple harmonic motion are observed to be 2 cm , 1 m/s , and 10 m/s ^2 , For a particle executing simple harmonic motion, the maximum speed and acceleration are 10 cm/s and 50 cm/s ^2 , respectively. Determine the position(s) of the particle when its spA particle having a mass of 10 g oscillates according to the equation x = (2.0 cm ) [(100 s ⁻¹) t + /6] . Determine the amplitude, the time period, and the spring constant for thisA particle undergoes simple harmonic motion having an amplitude of 10 cm . At what distance from the mean position will its kinetic and potential energies be equal?A particle executes simple harmonic motion with an amplitude of 10 cm and a time period of 6 s . At t = 0 , it is located at x = 5 cm and is moving towards the positive x-directionA particle having a mass of 10 g oscillates according to the equation x = (2.0 cm ) [(100 s ⁻¹) t + /6] . Determine the position, the velocity, and the acceleration of the particleA particle begins its motion at t = 0 according to the equation x = 5 (20 t + /3) , where x is in centimetre and t is in second. At what time will the particle first have zero acceA particle starts moving at t = 0 with its position given by the equation x = 5 (20 t + /3) , where x is in centimetre and t is in second. Determine the time when the particle firs Full Simple Harmonic Motion list All Concepts Of Physics MCQ Edition [Volume 1] PYQs