A spring executes S.H.M. with mass 1 kg attached to it. The force constant of the spring is 4 N/m. If at any instant its velocity is 20 cm/s, the displacement at that instant is (Amplitude of S.H.M. is 0.4 m) ______.
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Always ensure your units are perfectly matched before doing any algebra! Mixing $20 \text{ cm/s}$ with $0.4 \text{ m}$ amplitude is a guaranteed way to land on the wrong answer. Convert everything to SI base units first.
Step 1: Understanding the Concept:
The velocity of a particle in SHM is related to its displacement $x$ by $v = \omega \sqrt{A^2 - x^2}$. Step 2: Formula Application:
$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{4}{1}} = 2$ rad/s.
$v = 20$ cm/s $= 0.2$ m/s.
$A = 0.4$ m. Step 3: Explanation:
$0.2 = 2 \sqrt{(0.4)^2 - x^2}$
$0.1 = \sqrt{0.16 - x^2}$
Squaring both sides:
$0.01 = 0.16 - x^2 \implies x^2 = 0.15$.
$x = \sqrt{0.15}$ m. Step 4: Final Answer:
The displacement is $\sqrt{0.15}$ m.