
A spring executes SHM with mass of 10 kg attached to it. The force constant of spring is 10N/m. If at any instant its velocity is 40 cm/sec, the displacement will be (where amplitude is 0.5m)
A. 0.09 m
B. 0.3 m
C. 0.03 m
D. 0.9 m
Answer
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Hint:The question is based on the waves and oscillation section of physics. We need to apply the expression of velocity of a particle during simple harmonic motion. After putting the values of given quantity we can easily calculate the displacement of the particle.
Formula Used:
Velocity of particle in SHM:
$v = \omega \sqrt {{A^2} - {x^2}} $
Where $v$= velocity, $A$= amplitude, $\omega $ = angular frequency and $x$= displacement from mean position.
Complete step by step solution:
The equation for the velocity of a particle undergoing simple harmonic motion is given below.
$v = \omega \sqrt {{A^2} - {x^2}} $
The value of amplitude (A) = 0.5m and velocity (v) = 0.4 m/s are given in the problem.
We have to find $\omega $ using the equation $\omega = \sqrt {\frac{k}{m}} $.
The value of force constant (k) = 10 N/m and mass (m) 10 kg is given in the problem.
\[\omega = \sqrt {\frac{k}{m}} \\
\Rightarrow \omega = \sqrt {\frac{{10}}{{10}}} \\
\Rightarrow \omega = 1\,rad/s\]
Then, substitute the values and find the displacement from the velocity equation.
$v = \omega \sqrt {{A^2} - {x^2}} $
$\Rightarrow 0.4 = 1\sqrt {{{0.5}^2} - {x^2}} $
$\Rightarrow 0.16 = 0.25 - {x^2}$
$\Rightarrow {x^2} = 0.25 - 0.16$
$\therefore x = \sqrt {0.09} = 0.3\,m$
Hence, the correct option is option B.
Additional Information: Simple harmonic motion (SHM) is defined as an oscillatory motion where the particle's acceleration at any position is inversely proportional to its displacement from the mean position. All the SHM are oscillatory and periodic, but not all oscillatory motions are SHM.
Note: Understanding the fundamentals of the simple harmonic motion is crucial to understanding the properties of alternating currents, light waves, and sound waves. Simple Harmonic Motion can be classified into two types: linear SHM and angular SHM. There are two conditions of simple harmonic motion, first one is $F \propto - x$ and second one is $a \propto - x$, where F = restoring force, a = acceleration and x = displacement of the particle from the mean position.
Formula Used:
Velocity of particle in SHM:
$v = \omega \sqrt {{A^2} - {x^2}} $
Where $v$= velocity, $A$= amplitude, $\omega $ = angular frequency and $x$= displacement from mean position.
Complete step by step solution:
The equation for the velocity of a particle undergoing simple harmonic motion is given below.
$v = \omega \sqrt {{A^2} - {x^2}} $
The value of amplitude (A) = 0.5m and velocity (v) = 0.4 m/s are given in the problem.
We have to find $\omega $ using the equation $\omega = \sqrt {\frac{k}{m}} $.
The value of force constant (k) = 10 N/m and mass (m) 10 kg is given in the problem.
\[\omega = \sqrt {\frac{k}{m}} \\
\Rightarrow \omega = \sqrt {\frac{{10}}{{10}}} \\
\Rightarrow \omega = 1\,rad/s\]
Then, substitute the values and find the displacement from the velocity equation.
$v = \omega \sqrt {{A^2} - {x^2}} $
$\Rightarrow 0.4 = 1\sqrt {{{0.5}^2} - {x^2}} $
$\Rightarrow 0.16 = 0.25 - {x^2}$
$\Rightarrow {x^2} = 0.25 - 0.16$
$\therefore x = \sqrt {0.09} = 0.3\,m$
Hence, the correct option is option B.
Additional Information: Simple harmonic motion (SHM) is defined as an oscillatory motion where the particle's acceleration at any position is inversely proportional to its displacement from the mean position. All the SHM are oscillatory and periodic, but not all oscillatory motions are SHM.
Note: Understanding the fundamentals of the simple harmonic motion is crucial to understanding the properties of alternating currents, light waves, and sound waves. Simple Harmonic Motion can be classified into two types: linear SHM and angular SHM. There are two conditions of simple harmonic motion, first one is $F \propto - x$ and second one is $a \propto - x$, where F = restoring force, a = acceleration and x = displacement of the particle from the mean position.
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