Question 227: Oscillations

Question

A spring executes S.H.M. with mass 1 kg attached to it. The force constant of the spring is $4 \mathrm{~N} / \mathrm{m}$. If at any instant its velocity is $20 \mathrm{~cm} / \mathrm{s}$, the displacement at that instant is (Amplitude of S.H.M. is 0.4 m )

Accepted Answer

We are given:
Mass $m = 1 \, 	ext{kg}$
Spring constant $k = 4 \, 	ext{N/m}$
Velocity at some instant $v = 20 \, 	ext{cm/s} = 0.2 \, 	ext{m/s}$
Amplitude $A = 0.4 \, 	ext{m}$
We need the displacement $x$ at that instant.
Step 1: Angular frequency
$ \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{4}{1}} = 2 \, 	ext{rad/s} $
Step 2: Energy conservation in SHM
Total energy (constant):
$ E = 	frac{1}{2} k A^2 = 	frac{1}{2} m \omega^2 A^2 $
At displacement $x$ and velocity $v$:
$ E = 	frac{1}{2} m v^2 + 	frac{1}{2} k x^2 $
Step 3: Equate energies
$ 	frac{1}{2} m v^2 + 	frac{1}{2} k x^2 = 	frac{1}{2} k A^2 $
Substitute values:
$ 	frac{1}{2} (1)(0.2^2) + 	frac{1}{2}(4) x^2 = 	frac{1}{2} (4)(0.4^2) $
$ 0.5 	imes 0.04 + 2 x^2 = 2 	imes 0.16 $
$ 0.02 + 2x^2 = 0.32 $
$ 2x^2 = 0.30 $
$ x^2 = 0.15 $
$ x = \sqrt{0.15} \, 	ext{m} $
Ã¢ÂÂ Final Answer:
The displacement is
Option B: $\sqrt{0.15} \, 	ext{m}$

Answer

$\sqrt{0.11} \mathrm{~m}$

Answer

$\sqrt{0.15} \mathrm{~m}$

Answer

$\sqrt{0.17} \mathrm{~m}$

Answer

$\sqrt{0.19} \mathrm{~m}$

Oscillations

A spring executes S.H.M. with mass 1 kg attached to it. The force constant of the spring is 4 \mathrm{~N} / \mathrm{m}. If at any instant its velocity is 20 \mathrm{~cm} / \mathrm{s}, the displacement at that instant is (Amplitude of S.H.M. is 0.4 m )

Choose the correct answer: