Question 185: Oscillations

Question

A particle is doing simple harmonic motion of amplitude $$0.06 \mathrm{~m}$$ and time period $$3.14 \mathrm{~s}$$. The maximum velocity of the particle is _________ $$\mathrm{cm} / \mathrm{s}$$.

Accepted Answer

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 $v_{max}$:
$v_{max} = A\omega = 0.06 	imes \frac{2\pi}{3.14}$
$v_{max} = 0.06 	imes \frac{2 	imes 3.14}{3.14}$
$v_{max} = 0.06 	imes 2$
$v_{max} = 0.12 \, \mathrm{m/s}$
To convert meters per second to centimeters per second, we use the conversion factor $1 \, \mathrm{m/s} = 100 \, \mathrm{cm/s}$. Therefore,
$v_{max} = 0.12 \, \mathrm{m/s} 	imes 100 \, \mathrm{cm/m} = 12 \, \mathrm{cm/s}$
Thus, the maximum velocity of the particle is $12 \, \mathrm{cm/s}$.

Oscillations

A particle is doing simple harmonic motion of amplitude $0.06 \mathrm{~m} and time period 3.14 \mathrm{~s}. The maximum velocity of the particle is _________ \mathrm{cm} / \mathrm{s}$.

Choose the correct answer: