QJEEM24Ph6e5e: JEE Main 2024 Physics - Class 11 JEEM24Ph6e5e Solution: A particle is doing simple harmonic motion of amplitude 0.06…

For a particle performing simple harmonic motion (SHM), the maximum velocity $$v_{max}$$ can be calculated using the formula:

$v_{max} = A\omega$

where $A$ is the amplitude of the motion and $\omega$ is the angular frequency. The angular frequency $\omega$ is related to the time period $T$ by the formula:

$\omega = \frac{2\pi}{T}$

Given:

• Amplitude, $A = 0.06 \, \mathrm{m}$


• Time period, $T = 3.14 \, \mathrm{s}$

First, we find the angular frequency:

$\omega = \frac{2\pi}{T} = \frac{2\pi}{3.14}$

Substituting $\omega$ and $A$ in the formula for $v_{max}$ :

$v_{max} = A\omega = 0.06 \times \frac{2\pi}{3.14}$

$v_{max} = 0.06 \times \frac{2 \times 3.14}{3.14}$

$v_{max} = 0.06 \times 2$

$v_{max} = 0.12 \, \mathrm{m/s}$

To convert meters per second to centimeters per second, we use the conversion factor $1 \, \mathrm{m/s} = 100 \, \mathrm{cm/s}$ . Therefore,

$v_{max} = 0.12 \, \mathrm{m/s} \times 100 \, \mathrm{cm/m} = 12 \, \mathrm{cm/s}$

Thus, the maximum velocity of the particle is $12 \, \mathrm{cm/s}$ .