Question 258: Oscillations

Question

A mass suspended from a vertical spring performs S.H.M. of period 0.1 second. The spring is unstretched at the highest point of suspension. Maximum speed of the mass is (Gravitational acceleration $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )

Accepted Answer

Given:
Time for one complete swing (period) $T = 0.1$ seconds
The spring is
not stretched
when the mass is at its highest point (this is the farthest point from the middle position).
Gravitational acceleration, $g = 10\, \mathrm{m/s}^2$
Let's find the answer step by step.
Step 1: Formula for Maximum Speed in S.H.M.
In simple harmonic motion, the highest speed is given by:
$$ v_{max} = A \omega $$
Where:
$A$ is the amplitude (the maximum distance from the center position)
$\omega$ is the angular frequency (tells us how fast the object goes in a circle or swings back and forth)
Step 2: Find Angular Frequency $\omega$
The formula connecting period and angular frequency is:
$$ \omega = \frac{2\pi}{T} $$
Plug in $T = 0.1$:
$$ \omega = \frac{2\pi}{0.1} = 20\pi $$
Step 3: Find Amplitude $A$
At the highest position, the spring is just relaxed (not stretched). This point is $A$ distance from the mean (middle) position.
At the middle (mean position), gravity balances the pull of the spring: $mg = kx_0$, where $x_0$ is the stretch from natural length.
At highest point: extension is $0$ (spring just relaxed). This distance from the middle to the top is our amplitude $A$.
So, $A = x_0$.
Thus, to find $A$ we find $x_0$.
From $mg = kx_0$:
$$ x_0 = \frac{mg}{k} $$
Step 4: Connect $k$ and $m$ to $\omega$
For a spring and mass, angular frequency is:
$$ \omega = \sqrt{\frac{k}{m}} $$
This means $ \frac{k}{m} = \omega^2 $.
So, substitute into earlier expression for $x_0$:
$$ x_0 = \frac{mg}{k} = \frac{g}{k/m} = \frac{g}{\omega^2} $$
Step 5: Plug Amplitude into the Speed Formula
Recall:
$$ v_{max} = A\omega $$
Since $A = x_0 = \frac{g}{\omega^2}$, then:
$$ v_{max} = \frac{g}{\omega^2} \cdot \omega = \frac{g}{\omega} $$
Step 6: Substitute the Values
$g = 10$
$\omega = 20\pi$
Now calculate maximum speed:
$$ v_{max} = \frac{10}{20\pi} = \frac{1}{2\pi} \ \mathrm{m/s} $$
Step 7: Final Answer
The correct answer is:
$$ \boxed{\frac{1}{2\pi} \,\mathrm{m/s}} $$

Answer

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

Answer

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

Answer

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

Answer

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

Oscillations

A mass suspended from a vertical spring performs S.H.M. of period 0.1 second. The spring is unstretched at the highest point of suspension. Maximum speed of the mass is (Gravitational acceleration \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2 )

Choose the correct answer: