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Hint: The angular velocity \[\omega \] in terms of maximum velocity and acceleration is
\[\dfrac{{{a}_{\max }}}{{{v}_{\max }}}=\omega \]
Time period of S.H.M is inversely proportional to the angular velocity.
Formula used: If the maximum displacement (amplitude) of a particle exhibiting S.H.M. be \[\alpha \] and the angular velocity be \[\omega \]. Then maximum velocity is given by,
\[{{v}_{\max }}=\alpha \omega \]
And, the maximum acceleration is
\[{{a}_{\max }}=\alpha {{\omega }^{2}}\]
The time period of S.H.M is generally given by
\[T=\dfrac{2\pi }{\omega }\]
In terms of maximum velocity and acceleration,
\[\dfrac{{{a}_{\max }}}{{{v}_{\max }}}=\omega \]
So, the time period is
\[
T=\dfrac{2\pi }{{}^{{{a}_{\max }}}/{}_{{{v}_{\max }}}} \\
T=2\pi \cdot \dfrac{{{v}_{\max }}}{{{a}_{\max }}} \\
\]
Where T is the time period, \[{{v}_{\max }}\] is the maximum velocity of the particle and \[{{a}_{\max }}\] is the maximum acceleration of the particle.
Complete step by step solution:
The maximum velocity of the particle, \[{{v}_{\max }}=100\text{ cm/sec}\]
The maximum acceleration of the particle, \[{{a}_{\max }}=157\text{ cm/se}{{\text{c}}^{2}}\]
Substitute the value of maximum velocity and maximum acceleration in the time-period formula:
\[
T=2\pi \cdot \dfrac{100\text{ cms/sec}}{157\text{ cm/se}{{\text{c}}^{2}}} \\
\text{ }=2\pi \cdot 0.6369\text{ sec} \\
\text{ }=4\text{ sec} \\
\]
Therefore, the time period of the S.H.M is \[4\text{ sec}\].
Hence, the correct answer is option A.
Additional information:
When a particle moves in a straight line to and fro about its equilibrium position in such a way that its acceleration is always directly proportional to its displacement and directed towards the equilibrium position, then the motion of the particle is called simple harmonic motion.
In a graphical representation of displacement, velocity and acceleration in S.H.M.:
All three quantities vary harmonically with time, having the same time period.
The displacement curve in a sinusoidal curve.
The acceleration curve is a mirror image of the displacement curve.
Note: The velocity of the particle at the instant of passing through the mean position is maximum and is minimum at the extreme positions.
The acceleration of the particle is maximum at the extreme points and minimum at the equilibrium position.
\[\dfrac{{{a}_{\max }}}{{{v}_{\max }}}=\omega \]
Time period of S.H.M is inversely proportional to the angular velocity.
Formula used: If the maximum displacement (amplitude) of a particle exhibiting S.H.M. be \[\alpha \] and the angular velocity be \[\omega \]. Then maximum velocity is given by,
\[{{v}_{\max }}=\alpha \omega \]
And, the maximum acceleration is
\[{{a}_{\max }}=\alpha {{\omega }^{2}}\]
The time period of S.H.M is generally given by
\[T=\dfrac{2\pi }{\omega }\]
In terms of maximum velocity and acceleration,
\[\dfrac{{{a}_{\max }}}{{{v}_{\max }}}=\omega \]
So, the time period is
\[
T=\dfrac{2\pi }{{}^{{{a}_{\max }}}/{}_{{{v}_{\max }}}} \\
T=2\pi \cdot \dfrac{{{v}_{\max }}}{{{a}_{\max }}} \\
\]
Where T is the time period, \[{{v}_{\max }}\] is the maximum velocity of the particle and \[{{a}_{\max }}\] is the maximum acceleration of the particle.
Complete step by step solution:
The maximum velocity of the particle, \[{{v}_{\max }}=100\text{ cm/sec}\]
The maximum acceleration of the particle, \[{{a}_{\max }}=157\text{ cm/se}{{\text{c}}^{2}}\]
Substitute the value of maximum velocity and maximum acceleration in the time-period formula:
\[
T=2\pi \cdot \dfrac{100\text{ cms/sec}}{157\text{ cm/se}{{\text{c}}^{2}}} \\
\text{ }=2\pi \cdot 0.6369\text{ sec} \\
\text{ }=4\text{ sec} \\
\]
Therefore, the time period of the S.H.M is \[4\text{ sec}\].
Hence, the correct answer is option A.
Additional information:
When a particle moves in a straight line to and fro about its equilibrium position in such a way that its acceleration is always directly proportional to its displacement and directed towards the equilibrium position, then the motion of the particle is called simple harmonic motion.
In a graphical representation of displacement, velocity and acceleration in S.H.M.:
All three quantities vary harmonically with time, having the same time period.
The displacement curve in a sinusoidal curve.
The acceleration curve is a mirror image of the displacement curve.
Note: The velocity of the particle at the instant of passing through the mean position is maximum and is minimum at the extreme positions.
The acceleration of the particle is maximum at the extreme points and minimum at the equilibrium position.
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