If \(x=5\sin(\pi i+\frac{\pi}{3})m\) represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are :

Question

The variation of displacement with time of a simple harmonic motion (SHM) for a particle of mass \( m \) is represented by: \[ y = 2 \sin \left( \frac{\pi}{2} + \phi \right) \, \text{cm} \] The maximum acceleration of the particle is:

Answer 1

Concepts: Simple harmonic motion, Amplitude, Time period

Explanation:

The provided equation, x = 5sin(πt + π/3) m, describes simple harmonic motion (SHM). This equation is in the standard form x = A sin(ωt + φ), where A represents the amplitude, ω represents the angular frequency, and φ represents the phase constant.

From the equation, we can determine:

Amplitude (A): This is the coefficient of the sine function, which is 5 meters.

Angular frequency (ω): This is the coefficient of t within the sine function, which is π.

Time period (T): The time period of SHM is calculated using the formula T = 2π/ω. Substituting ω = π, we get T = 2π/π = 2 seconds.

Therefore, the amplitude is 5 meters and the time period is 2 seconds.

Step by Step Solution:

Step 1: Determine the amplitude A from the SHM equation x = A sin(ωt + φ). In this case, A = 5 meters.

Step 2: Identify the angular frequency ω from the SHM equation. Here, ω = π.

Step 3: Apply the formula for the time period: T = 2π/ω.

Step 4: Substitute ω = π into the formula to calculate T: T = 2π/π = 2 seconds.

Final Answer:

The amplitude is 5 meters and the time period is 2 seconds.

If \(x=5\sin(\pi i+\frac{\pi}{3})m\) represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are :

The Correct Option is B

Solution and Explanation

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