Question:medium

A child is sitting on a swing which performs S.H.M. It has minimum and maximum heights from the ground as $0.75\text{ m}$ and $2\text{ m}$ respectively. Its maximum speed will be [$g = 10\text{ m/s}^2$]

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The maximum velocity calculation for an oscillating swing or pendulum is identical to finding the final velocity of an object falling freely from a height of $\Delta h$. Using the standard kinematic shortcut $v = \sqrt{2g\Delta h}$ bypasses S.H.M. notation completely.
Updated On: Jun 4, 2026
  • $1.25\text{ m/s}$
  • $12.5\text{ m/s}$
  • $5\text{ m/s}$
  • $25\text{ m/s}$
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Picture the swing.
A child on a swing moves like a pendulum. At the highest point the swing stops for a moment, and at the lowest point it moves fastest. The lowest height is $0.75$ m and the highest height is $2$ m above the ground.
Step 2: Use energy idea.
As the swing drops from the top to the bottom, its stored height energy turns into motion energy. Nothing is lost, so the drop in potential energy equals the gain in kinetic energy.
Step 3: Write the energy balance.
\[ \frac{1}{2}mv_{\max}^2 = mg\,(h_{\max} - h_{\min}) \] The mass $m$ is on both sides, so it cancels.
Step 4: Solve for the speed.
\[ v_{\max} = \sqrt{2g\,(h_{\max} - h_{\min})} \]
Step 5: Find the height drop.
\[ h_{\max} - h_{\min} = 2 - 0.75 = 1.25\ \text{m} \]
Step 6: Put in the numbers.
With $g = 10$ m/s$^2$: \[ v_{\max} = \sqrt{2 \times 10 \times 1.25} = \sqrt{25} = 5\ \text{m/s} \] So the greatest speed of the swing is $5$ m/s, which is option (3). \[ \boxed{v_{\max} = 5\ \text{m/s}} \]
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