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Simple Harmonic Motion Test - 9

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Simple Harmonic Motion Test - 9
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  • Question 1
    1 / -0

    Two simple harmonic motions of angular frequencies 100 and 1000 rad s-1 have the same displacement amplitude. The ratio of their maximum acceleration is:

    Solution

     

  • Question 2
    1 / -0

    Two points are located at a distance of 10 m and 15 m from the source of oscillation. The period of oscillation is 0.05 s and the velocity of the wave is 300 m/s. What is the phase difference between the oscillations of two points?

    Solution

    Hint: The phase difference depends on the path difference.

    Step 1: Identify the given values.

    The path difference, Δy=5 m

    Time period of oscillations, T = 0.05 sec

    Velocity of the wave, v = 300 m/s

    Step 2: Find the wavelength of the wave.
    Now,

    Step 3: Find the phase difference.

    The phase difference ϕ is given as: ϕ = 2π/λ × (Path difference)

     

  • Question 3
    1 / -0

    A particle executes simple harmonic oscillation with an amplitude a. The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is:

    Solution

    Hint: The displacement of the particle will be a2a2.

    Step 1: Find the displacement equation.

    Let displacement equation of particle is,

    y= asinωt

    Step 2: Put the value of the displacement.
    As y = a/2,
    Therefore, a/2 = asinωt
    Or, sinωt = 1/2 = sin π/6
    Or, ωt = π/6
    Or, t = π/6ω

    Step 3: Put the value of ω.

     

  • Question 4
    1 / -0

    The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion is:

    Solution

    Hint: The acceleration is given by, a = dv/dt

    Step 1: Find the velocity and acceleration in SHM.

    The displacement equation of a particle executing SHM is, 
    x = acos(ωt +ϕ)                            …….(i) 
    Velocity, v = -aωsin(ωt +ϕ) = aωcos(ωt +ϕ+π/2)          …….(ii) 

    Step 2: Find the phase difference.

    The phase difference between the velocity and the acceleration is π/2.

     

  • Question 5
    1 / -0

    A rectangular block of mass m and area of cross-section A floats in a liquid of density ρ. If it is given a small vertical displacement from equilibrium, it undergoes oscillation with a time period T. Then:

    Solution

    Hint: The upthrust applied on the block will be the restoring force.

    Step 2: Find upthrust on the block.

    Let block is displaced through x from mean position,

    Then the weight of displaced water or upthrust (upwards)  = -Axρg 
    where A is an area of the cross-section of the block and ρ is its density.

    Step 2: Find the acceleration of the block.

    This upthrust provides a restoring force to the block.

    So,  ma = -Axρg

    This is the equation of simple harmonic motion.

    Step 3: Find the time period of oscillations of the block.
    The time period of oscillation,

     

  • Question 6
    1 / -0

    The displacement of a particle executing simple harmonic motion is given by,

    y = A+ A sinωt + B cosωt.

    Then the amplitude of its oscillation is given by:

    Solution

    Hint: Use the principle of superposition.

    Step 1: Find the combined equation of oscillation.

     

  • Question 7
    1 / -0

    The average velocity of a particle executing SHM in one complete vibration is:

    Solution

    Hint: The initial and final positions of the particle are the same.

    Step 1: Find the displacement of the particle.

    The displacement of the particle is zero.

    Step 2: Find the average velocity.

    The average velocity of a particle executing SHM in one complete vibration, vavg = 0

     

  • Question 8
    1 / -0

    Two pendulums have time periods T and 5T/4. They start SHM at the same time from the mean position. What will be the phase difference between them after the bigger pendulum completed one oscillation?

    Solution

    When a bigger pendulum of time period (5T/4) completes one oscillation, the smaller pendulum will complete (5/4) vibrations. It means the smaller pendulum will be leading the bigger pendulum by phase T/4 sec = π/2 rad = 90°

     

  • Question 9
    1 / -0

    A simple harmonic oscillator has amplitude 'a' and time period T. The time required by it to travel from x=a to x = a/2 is

    Solution

    It is required to calculate the time for extreme positions. Hence, in this case, the equation of displacement of the particle can be written as 

     

  • Question 10
    1 / -0

    The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad/min, what is the maximum speed?

    Solution

    Given the angular frequency of the piston, ω = 200 rad/min

    Stroke length=1m

     

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