A particle of mass 0.1 kg0.1 kg is executing simple harmonic motion of amplitude 0.1 m0.1 m. When the particle passes through the mean position, its kinetic energy is 8×10−3 J8×10-3 J. If the initial phase is 45°,45°, the equation of its motion is (Assume, x (t)𝑥 𝑡 as the position of the particle at time t𝑡)
Question
A particle of mass 0.1 kg0.1 kg is executing simple harmonic motion of amplitude 0.1 m0.1 m. When the particle passes through the mean position, its kinetic energy is 8×10−3 J8×10-3 J. If the initial phase is 45°,45°, the equation of its motion is (Assume, x (t)𝑥 𝑡 as the position of the particle at time t𝑡)
Solution
The equation of motion for a simple harmonic oscillator is given by:
x(t) = A cos(ωt + φ)
where:
- x(t) is the position of the particle at time t,
- A is the amplitude of the motion,
- ω is the angular frequency,
- φ is the phase constant.
Given in the problem, we have:
- A = 0.1 m (amplitude),
- φ = 45° (initial phase).
We need to find the angular frequency ω. We know that the kinetic energy (K.E.) of a simple harmonic oscillator when it passes through the mean position is given by:
K.E. = 1/2 mω²A²
where m is the mass of the particle. Substituting the given values:
8×10^-3 J = 1/2 * 0.1 kg * ω² * (0.1 m)²
Solving this equation for ω, we get:
ω = sqrt((2 * 8×10^-3 J) / (0.1 kg * (0.1 m)²)) = 40 rad/s
Therefore, the equation of motion for the particle is:
x(t) = 0.1 m * cos(40t + 45°)
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