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A particle is attached to a vertical spr...

A particle is attached to a vertical spring and is pulled down a distance 0.04m below its equilibrium position and is released from rest. The initial upward acceleration of the particle is `0.30 ms^(-2)`. The period of the oscillation is

A

4.08 s

B

1.92 s

C

3.90 s

D

2.29 s

Text Solution

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The correct Answer is:
To solve the problem step by step, we will follow the reasoning laid out in the video transcript. ### Step 1: Understanding the Problem A particle is attached to a vertical spring and is pulled down 0.04 m from its equilibrium position and released. The initial upward acceleration is given as 0.30 m/s². We need to find the period of oscillation of the particle. ### Step 2: Relating Acceleration to Angular Frequency The acceleration \( a \) of a particle in simple harmonic motion can be expressed as: \[ a = -\omega^2 x \] where \( \omega \) is the angular frequency and \( x \) is the displacement from the equilibrium position. Since we are interested in the magnitude: \[ a = \omega^2 x \] ### Step 3: Substituting Known Values From the problem, we know: - \( a = 0.30 \, \text{m/s}^2 \) - \( x = 0.04 \, \text{m} \) Substituting these values into the equation: \[ 0.30 = \omega^2 \cdot 0.04 \] ### Step 4: Solving for Angular Frequency \( \omega \) Rearranging the equation to solve for \( \omega^2 \): \[ \omega^2 = \frac{0.30}{0.04} \] Calculating the right side: \[ \omega^2 = \frac{0.30 \times 100}{0.04 \times 100} = \frac{30}{4} = 7.5 \] Now, taking the square root to find \( \omega \): \[ \omega = \sqrt{7.5} \approx \frac{\sqrt{30}}{2} \] ### Step 5: Finding the Time Period \( T \) The period \( T \) of oscillation is given by the formula: \[ T = \frac{2\pi}{\omega} \] Substituting the value of \( \omega \): \[ T = \frac{2\pi}{\sqrt{7.5}} = \frac{2\pi}{\sqrt{30}/2} = \frac{4\pi}{\sqrt{30}} \] ### Step 6: Calculating the Numerical Value Now we calculate \( T \): \[ T = \frac{4 \times 3.14}{\sqrt{30}} \approx \frac{12.56}{5.477} \approx 2.29 \, \text{s} \] ### Final Answer Thus, the period of oscillation is approximately: \[ \boxed{2.29 \, \text{s}} \]
To solve the problem step by step, we will follow the reasoning laid out in the video transcript. ### Step 1: Understanding the Problem A particle is attached to a vertical spring and is pulled down 0.04 m from its equilibrium position and released. The initial upward acceleration is given as 0.30 m/s². We need to find the period of oscillation of the particle. ### Step 2: Relating Acceleration to Angular Frequency The acceleration \( a \) of a particle in simple harmonic motion can be expressed as: \[ ...
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