If x = 5sin (πt +π/3) m m represents the motion of a particle executing simple harmonic motion, the amplitude and respectively, are:

Physics Problem: Amplitude and Time Period of Simple Harmonic Motion

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

• (a) \( 5 \, \text{cm}, 1 \, \text{s} \)
• (b) \( 5 \, \text{m}, 1 \, \text{s} \)
• (c) \( 5 \, \text{cm}, 2 \, \text{s} \)
• (d) \( 5 \, \text{m}, 2 \, \text{s} \)

Detailed Solution

The general equation for simple harmonic motion is:

\[ x = A \sin(\omega t + \phi) \]

where:

• \( A \) is the amplitude,
• \( \omega \) is the angular frequency,
• \( \phi \) is the phase constant.

Given the equation:

\[ x = 5 \sin(\pi t + \phi) \, \text{m} \]

We can compare it with the general equation to determine:

\[ A = 5 \, \text{m} \quad \text{and} \quad \omega = \pi \, \text{rad/s} \]

The time period \( T \) is related to the angular frequency \( \omega \) by:

\[ T = \frac{2\pi}{\omega} \]

Substituting \( \omega = \pi \):

\[ T = \frac{2\pi}{\pi} = 2 \, \text{s} \]

Thus, the amplitude is \( 5 \, \text{m} \), and the time period is \( 2 \, \text{s} \).

Answer: (d) \( 5 \, \text{m}, 2 \, \text{s} \)