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, $6\pi \, \text{m/s}$.