A particle is executing SHM. If the displacement at any instant is given by x = 3 c o s 2 + 4 s in 2 t. What is the time period of the particle?a)1.57 Sb)2 Sc)3.14 Sd)5 SCorrect answer is option 'C'. Can you explain this answer?

Question

Answer

x = 3 cos 2t + 4 sin 2t
x = 3 cos 2t + 4 sin 2t

⇒ ω = 2

Answer

Given:
The displacement of a particle executing Simple Harmonic Motion (SHM) is given by the equation: x = 3cos(2πt) + 4sin(2πt)

To Find:
The time period of the particle.

Explanation:
1. Equation of SHM:
The general equation for SHM is given by:
x = A cos(ωt + φ)

where,
x = displacement of the particle
A = amplitude of the motion
ω = angular frequency of the motion
t = time
φ = phase constant

In the given equation, x = 3cos(2πt) + 4sin(2πt), we can rewrite it by combining the sine and cosine terms:
x = A cos(ωt + φ)

Comparing the terms, we can see that:
A = √(3^2 + 4^2) = 5
ωt + φ = 2πt

2. Relationship between Time Period and Angular Frequency:
The time period (T) is the time taken to complete one full oscillation. The angular frequency (ω) is related to the time period by the equation:
T = 2π/ω

3. Finding the Time Period:
From the equation, ωt + φ = 2πt, we can equate the coefficients of t:
ωt = 2πt

This implies that ω = 2π, which is the angular frequency of the particle.

Using the relationship between time period and angular frequency, we can find the time period:
T = 2π/ω
T = 2π/(2π)
T = 1

4. Answer:
The time period of the particle is 1 second, which corresponds to option (b) in the given choices.

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