Question 289: Oscillations

Question

Frequency of a particle performing S.H.M. is 10 Hz . The particle is suspended from a vertical spring. At the highest point of its oscillation the spring is unstretched. Maximum speed of the particle is $\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right)$

Accepted Answer

To find the maximum speed of the particle in simple harmonic motion (SHM) with frequency $10 \, 	ext{Hz}$, we will use the relationship between angular frequency, frequency, and maximum speed.
Determine Angular Frequency ($\omega$):
The angular frequency $\omega$ is related to the frequency $f$ by the formula:
$$ \omega = 2\pi f $$
Given $ f = 10 \, 	ext{Hz} $, we have:
$$ \omega = 2\pi 	imes 10 = 20\pi \, 	ext{rad/s} $$
Maximum Speed Formula in SHM:
The maximum speed $v_{	ext{max}}$ of a particle in SHM is given by:
$$ v_{	ext{max}} = A \omega $$
where $A$ is the amplitude of the oscillation.
Determine Amplitude ($A$):
At the highest point, the spring is unstretched, meaning the elongation due to the particle's weight is equal to the amplitude of oscillation.
Use the equilibrium condition $mg = kA$, and since $ \omega = \sqrt{\frac{k}{m}} $, $k = m\omega^2$.
Therefore, at maximum elongation, the net force provides the centripetal force for the circular motion:
$$ mg = m\omega^2 A $$
Consequently, $ A = \frac{g}{\omega^2} $.
Substitute Values to Find $v_{	ext{max}}$:
Substituting for amplitude:
$$ A = \frac{10}{(20\pi)^2} = \frac{10}{400\pi^2} = \frac{1}{40\pi^2} $$
Now, substitute $A$ and $\omega$ into $v_{	ext{max}} = A \omega$:
$$ v_{	ext{max}} = \left(\frac{1}{40\pi^2}ight) 	imes 20\pi = \frac{20\pi}{40\pi^2} $$
$$ v_{	ext{max}} = \frac{1}{2\pi} \, 	ext{m/s} $$
Thus, the maximum speed of the particle is $\frac{1}{2\pi} \, 	ext{m/s}$.
Answer:
Option B: $\frac{1}{2 \pi} \, 	ext{m/s}$

Answer

$\frac{1}{\pi} \mathrm{~m} / \mathrm{s}$

Answer

$\frac{1}{2 \pi} \mathrm{~m} / \mathrm{s}$

Answer

$\frac{1}{4 \pi} \mathrm{~m} / \mathrm{s}$

Answer

$2 \pi \mathrm{~m} / \mathrm{s}$

Oscillations

Frequency of a particle performing S.H.M. is 10 Hz . The particle is suspended from a vertical spring. At the highest point of its oscillation the spring is unstretched. Maximum speed of the particle is \left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right)

Choose the correct answer: