Question

The displacement of a particle executing SHM is given by X = 3 sin [2πt + π/4] , where 'X' is in meter and 't' is in second. The amplitude and maximum speed of the particle is 

• A. 3 m, 6π ms^-1
• B. 3 m, 2π ms^-1
• C. 3 m, 8π ms^-1
• D. 3 m, 4π ms^-1

Answer 1

The Correct Option is A

To find the amplitude and maximum speed of the particle, we are given the equation for displacement in simple harmonic motion (SHM):

\( X = 3 \sin(2\pi t + \frac{\pi}{4}) \)

Step 1: Compare with the general form of SHM:

The general equation for SHM is:

\( X = A \sin(\omega t + \phi) \)

By comparing the given equation with the general form, we can identify the following parameters:

• Amplitude (A): The coefficient of the sine function is 3, so \( A = 3 \, \text{m} \).
• Angular Frequency (ω): The coefficient of \( t \) inside the sine function is \( 2\pi \), so \( \omega = 2\pi \, \text{rad/s} \).

Step 2: Calculate the Maximum Speed (Vmax):

The maximum speed in SHM occurs when the displacement is at its maximum, which is equal to the amplitude \( A \). The maximum speed can be calculated using the formula:

\( V_{\text{max}} = \omega A \)

Substituting the values we found:

\( V_{\text{max}} = (2\pi \, \text{rad/s}) \times (3 \, \text{m}) = 6\pi \, \text{m/s} \)

Therefore, the amplitude and maximum speed of the particle are: 3 m and \( 6\pi \, \text{m/s} \), respectively. The correct option is (A) 3 m, \( 6\pi \, \text{m/s} \).

Answer 2

The displacement of a particle executing Simple Harmonic Motion (SHM) is given by the equation: \[ X(t) = A \sin(\omega t + \phi) \] where:

• \( X(t) \) is the displacement at time \( t \)
• \( A \) is the amplitude
• \( \omega \) is the angular frequency
• \( \phi \) is the phase constant

The given equation is: \[ X = 3 \sin \left(2\pi t + \frac{\pi}{4}\right) \] Here, \( X \) is in meters and \( t \) is in seconds.

By comparing the given equation with the standard SHM equation, we can identify the parameters:

• Amplitude: \( \mathbf{A = 3 \text{ m}} \)
• Angular frequency: \( \mathbf{\omega = 2\pi \text{ rad/s}} \)
• Phase constant: \( \phi = \frac{\pi}{4} \text{ rad} \)

The velocity \( V(t) \) of the particle is the time derivative of the displacement \( X(t) \): \[ V(t) = \frac{dX}{dt} = \frac{d}{dt} [A \sin(\omega t + \phi)] \] \[ V(t) = A\omega \cos(\omega t + \phi) \]

The speed of the particle is the magnitude of the velocity. The maximum speed \( V_{max} \) occurs when the cosine term is equal to \( \pm 1 \). \[ \mathbf{V_{max} = A\omega} \]

Substitute the values of \( A \) and \( \omega \) we found: \[ V_{max} = (3 \text{ m}) \times (2\pi \text{ rad/s}) \] \[ \mathbf{V_{max} = 6\pi \text{ m/s}} \]

Therefore, the amplitude of the particle is 3 m and the maximum speed is 6π m/s.

Comparing this result with the given options:

• 3 m, 6π ms^-1
• 3 m, 2π ms^-1
• 3 m, 8π ms^-1
• 3 m, 4π ms^-1

The correct option is the first one.