A particle performing simple harmonic motion is such that its amplitude is 4 m and speed of particle at mean position is 10 m/s. Find the distance of particle from mean position where velocity becomes 5 m/s.
- \(\sqrt3\; m\)
- \(2\sqrt3\; m\)
- \(\frac{\sqrt3}{2}\; m\)
- \(\frac{1}{\sqrt2}\; m\)
The Correct Option is B
Solution and Explanation
To solve this problem, we need to use the basic equations of simple harmonic motion (SHM). The velocity \(v\) of a particle in SHM at a distance \(x\) from the mean position is given by:
\(v = \omega \sqrt{A^2 - x^2}\)
where \(A\) is the amplitude and \(\omega\) is the angular frequency.
First, we compute the angular frequency \(\omega\) using the speed of the particle at the mean position. At the mean position, \(x = 0\), and the velocity \(v_{max} = 10 \, \text{m/s}\) is the maximum. Thus:
\(v_{max} = \omega A\)
Substitute the given values:
\(10 = \omega \times 4\)
Solve for \(\omega\):
\(\omega = \frac{10}{4} = 2.5 \, \text{rad/s}\)
Next, we find the distance \(x\) when the velocity \(v = 5 \, \text{m/s}\):
\(5 = 2.5 \sqrt{4^2 - x^2}\)
Simplifying this gives:
\(5 = 2.5 \sqrt{16 - x^2}\)
Divide both sides by 2.5:
\(2 = \sqrt{16 - x^2}\)
Square both sides to remove the square root:
\(4 = 16 - x^2\)
Rearrange to solve for \(x^2\):
\(x^2 = 16 - 4 = 12\)
Hence, solving for \(x\):
\(x = \sqrt{12} = 2\sqrt{3} \, \text{m}\)
Therefore, the distance of the particle from the mean position where the velocity becomes 5 m/s is \(2\sqrt{3} \; m\), matching the correct option.
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Concepts Used:
Simple Harmonic Motion
Simple Harmonic Motion is one of the most simple forms of oscillatory motion that occurs frequently in nature. The quantity of force acting on a particle in SHM is exactly proportional to the displacement of the particle from the equilibrium location. It is given by F = -kx, where k is the force constant and the negative sign indicates that force resists growth in x.
This force is known as the restoring force, and it pulls the particle back to its equilibrium position as opposing displacement increases. N/m is the SI unit of Force.
Types of Simple Harmonic Motion
Linear Simple Harmonic Motion:
When a particle moves to and fro about a fixed point (called equilibrium position) along with a straight line then its motion is called linear Simple Harmonic Motion. For Example spring-mass system
Conditions:
The restoring force or acceleration acting on the particle should always be proportional to the displacement of the particle and directed towards the equilibrium position.
- – displacement of particle from equilibrium position.
- – Restoring force
- - acceleration
Angular Simple Harmonic Motion:
When a system oscillates angular long with respect to a fixed axis then its motion is called angular simple harmonic motion.
Conditions:
The restoring torque (or) Angular acceleration acting on the particle should always be proportional to the angular displacement of the particle and directed towards the equilibrium position.
Τ ∝ θ or α ∝ θ
Where,
- Τ – Torque
- α angular acceleration
- θ – angular displacement