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: Understanding the Concept:
In SHM the speed at displacement \(x\) is \(v = \omega\sqrt{A^2 - x^2}\), where \(\omega = \sqrt{k/m}\).

Step 2: Find \(\omega\).
\(\omega = \sqrt{\dfrac{10}{10}} = 1\) rad/s.

Step 3: Use the speed.
\(v = 40\text{ cm/s} = 0.4\text{ m/s}\) and \(A = 0.5\) m.
\[ 0.4^2 = 1^2(0.5^2 - x^2) \Rightarrow 0.16 = 0.25 - x^2 \Rightarrow x^2 = 0.09 \Rightarrow x = 0.3\text{ m} \]

Step 4: Check the options.
\(x = 0.5\) m would be the extreme position where the speed is zero. \(x = 0.6\) m is bigger than the amplitude, which is impossible. \(x = 0.1\) m gives \(v = \sqrt{0.24} \approx 0.49\) m/s, not 0.4.

Final Answer:
The displacement is 0.3 m, option (B). \[ \boxed{0.3\text{ m}} \]