Question

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) ______.

• A. $\sqrt{0.11}$ m
• B. $\sqrt{0.15}$ m
• C. $\sqrt{0.17}$ m
• D. $\sqrt{0.19}$ m

Hint:

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.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given the mass, spring constant, amplitude, and an instantaneous velocity of an SHM system. We must calculate the exact displacement ($x$) occurring at that specific instant.

Step 2: Detailed Explanation:
The general velocity equation for a particle undergoing Simple Harmonic Motion (SHM) is:
$v = \omega \sqrt{A^2 - x^2}$
where:
$v$ = instantaneous velocity
$\omega$ = angular frequency
$A$ = Amplitude
$x$ = displacement
1. Find the angular frequency ($\omega$):
For a spring-mass system, the angular frequency is determined by:
$\omega = \sqrt{\frac{k}{m}}$
Given spring constant $k = 4 \text{ N/m}$ and mass $m = 1 \text{ kg}$.
$\omega = \sqrt{\frac{4}{1}} = 2 \text{ rad/s}$.
2. Standardize the units:
The velocity is given as $v = 20 \text{ cm/s}$.
Convert this to meters per second to match the amplitude and $\omega$:
$v = 0.2 \text{ m/s}$.
The amplitude is $A = 0.4 \text{ m}$.
3. Calculate the displacement ($x$):
Substitute the values into the velocity equation:
$0.2 = 2 \sqrt{(0.4)^2 - x^2}$
Divide by 2:
$0.1 = \sqrt{0.16 - x^2}$
Square both sides to remove the radical:
$(0.1)^2 = 0.16 - x^2$
$0.01 = 0.16 - x^2$
Rearrange to solve for $x^2$:
$x^2 = 0.16 - 0.01$
$x^2 = 0.15$
Take the square root:
$x = \sqrt{0.15} \text{ m}$

Step 3: Final Answer:
The displacement is $\sqrt{0.15}$ m, matching option (b).