Question

A spring executes S.H.M. with mass 10 kg attached to it. The force constant of spring is 10 N/m. If at any instant its velocity is 40 cm/s, the displacement at that instant is
(Amplitude of S.H.M. is 0.5 m)

• A. \(0.1\) m
• B. \(0.3\) m
• C. \(0.5\) m
• D. \(0.6\) m

Hint:

Find omega from the mass and force constant, then use v squared = omega squared times (A squared minus x squared).

Answer 1

The Correct Option is B

Step 1: Use energy.
Total energy $\dfrac{1}{2}kA^2 = \dfrac{1}{2}kx^2 + \dfrac{1}{2}mv^2$.

Step 2: Substitute.
$10(0.25) = 10x^2 + 10(0.16)$, since $m = 10$ kg and $v^2 = 0.16$.

Step 3: Solve.
$2.5 = 10x^2 + 1.6$, so $x^2 = 0.09$ and $x = 0.3$ m.

Final Answer:
Option (B). \[ \boxed{0.3\text{ m}} \]