
The kinetic energy of a particle executing SHM is $\;16J$ , when it is at its mean position. If the mass of the particle is $\;0.32kg$ , then what is the maximum velocity of the particle?
(A) $5m/s$
(B) $15m/s$
(C) $10m/s$
(D) $20m/s$
Answer
216.3k+ views
Hint: Here we have to use the concepts of simple harmonic motion. Simple harmonic motion can be defined as an oscillatory motion in which the acceleration of the particle at any point is directly proportional to the displacement of the mean position.
Complete step-by-step solution:
The acceleration is the variation in velocity with respect to time. If the velocity of the simple harmonic motion is maximum, the acceleration must be equal to zero. The acceleration is equal to zero only when the particle or object is at the initial position or if the displacement of the particle is zero. We know that the displacement of a particle is zero when the particle doesn’t change its initial position or it comes to the initial position after a certain time period.
Therefore, the particle will have a maximum velocity at the central position and minimum at the extreme positions.
At mean positions, the Kinetic energy ( K.E.) is given by,
$KE = \dfrac{1}{2}m.{v_{{{\max }^2}}}$
Here, $m$ = mass of the particle,
${v_{\max }}$ = velocity of the particle at its mean position
The given values in question are,
Mass of object $(m) = 0.32kg$
K.E. of object = $\;16J$
So here we put these values in the given formula to calculate the velocity of the object:
$16 = \dfrac{1}{2} \times 0.32 \times {v_{{{\max }^2}}}$
${v_{{{\max }^2}}} = \dfrac{{32}}{{0.32}}$
On further solving the above equation we get,
${v_{{{\max }^2}}} = 100$
Take the square root on both sides,
${v_{\max }} = \sqrt {100} $
${v_{\max }} = 10m/s$
So the maximum velocity of the particle at its mean position is $10m/s$ .
Hence, the correct answer is option (C) $10m/s$ .
Additional Information:
Not all oscillatory motions are simple harmonic whereas all harmonic motions are periodic and oscillatory. Oscillatory motion is often referred to as the harmonic motion of all oscillatory movements, the most important of which is basic harmonic motion.
In a simple harmonic motion, the coming back force or acceleration acting on the particle should always be equal to the displacement of the particle and be oriented towards the equilibrium state.
In other words, the same equation states the position of an object facing basic harmonic motion and to one component of the position of an object facing uniform circular motion.
In basic harmonic motion, the speed and displacement of the target are zero at an extreme location.
Note: Kinetic energy is zero at the extreme position as velocity is zero and potential energy is maximum. Potential energy is minimum at the mean position which is also the lowest position. If the motion of the pendulum makes a very small angle, then this motion can be approximated to Simple harmonic motion. One will need this information to solve SHM and pendulum involving questions.
Complete step-by-step solution:
The acceleration is the variation in velocity with respect to time. If the velocity of the simple harmonic motion is maximum, the acceleration must be equal to zero. The acceleration is equal to zero only when the particle or object is at the initial position or if the displacement of the particle is zero. We know that the displacement of a particle is zero when the particle doesn’t change its initial position or it comes to the initial position after a certain time period.
Therefore, the particle will have a maximum velocity at the central position and minimum at the extreme positions.
At mean positions, the Kinetic energy ( K.E.) is given by,
$KE = \dfrac{1}{2}m.{v_{{{\max }^2}}}$
Here, $m$ = mass of the particle,
${v_{\max }}$ = velocity of the particle at its mean position
The given values in question are,
Mass of object $(m) = 0.32kg$
K.E. of object = $\;16J$
So here we put these values in the given formula to calculate the velocity of the object:
$16 = \dfrac{1}{2} \times 0.32 \times {v_{{{\max }^2}}}$
${v_{{{\max }^2}}} = \dfrac{{32}}{{0.32}}$
On further solving the above equation we get,
${v_{{{\max }^2}}} = 100$
Take the square root on both sides,
${v_{\max }} = \sqrt {100} $
${v_{\max }} = 10m/s$
So the maximum velocity of the particle at its mean position is $10m/s$ .
Hence, the correct answer is option (C) $10m/s$ .
Additional Information:
Not all oscillatory motions are simple harmonic whereas all harmonic motions are periodic and oscillatory. Oscillatory motion is often referred to as the harmonic motion of all oscillatory movements, the most important of which is basic harmonic motion.
In a simple harmonic motion, the coming back force or acceleration acting on the particle should always be equal to the displacement of the particle and be oriented towards the equilibrium state.
In other words, the same equation states the position of an object facing basic harmonic motion and to one component of the position of an object facing uniform circular motion.
In basic harmonic motion, the speed and displacement of the target are zero at an extreme location.
Note: Kinetic energy is zero at the extreme position as velocity is zero and potential energy is maximum. Potential energy is minimum at the mean position which is also the lowest position. If the motion of the pendulum makes a very small angle, then this motion can be approximated to Simple harmonic motion. One will need this information to solve SHM and pendulum involving questions.
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