Question

A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time (in seconds) is
1. \(\frac{{4\pi }}{3}\)
2. \(\frac{{8\pi }}{3}\)
3. \(\frac{{7\pi }}{3}\)
4. \(\frac{{3\pi }}{8}\)

Answer 1

Correct Answer - Option 2 : \(\frac{{8\pi }}{3}\)

Concept:

The magnitude of its velocity in SI units is equal to that of its acceleration

In simple harmonic motion, position (x), velocity (v) and acceleration (α) of the particle are given by

Position (x) is given by,

x = A sin ωt

Velocity (v) is given by,

\(v = \omega \sqrt {{A^2} - {x^2}} \)

\(v = \omega \sqrt {{A^2} - {{\left( {A\sin \omega t} \right)}^2}}\) 

\(v = \omega \sqrt {{A^2} - {A^2}{{\sin }^2}\omega t}\) 

\(v = A\omega \sqrt {1 - {{\sin }^2}\omega t} \)

\(v = A\omega \sqrt {{{\cos }^2}\omega t}\) 

\(v = A\omega \cos \omega t\) 

Acceleration (\(\alpha \)) is given by,

α = -ω² x

α = -ω² A sin ωt

Calculation:

Given,

Amplitude = 5 cm.

Displacement x = 4 cm

At this time (when x = 4 cm), velocity and acceleration have same magnitude.

\(\Rightarrow \left| {{v_{x = 4}}} \right| = \left| {{\alpha _{x = 4}}} \right|\;\)

\(\left| {\omega \sqrt {{A^2} - {x^2}} } \right| = \left| { - {\omega ^2}x} \right|\)

\(\left| {\omega \sqrt {{5^2} - {4^2}} } \right| = \left| { - 4{\omega ^2}} \right|\)

\(\left| {\omega \sqrt {25 - 16} } \right| = \left| { - 4{\omega ^2}} \right|\)

\(\left| {\omega \sqrt 9 } \right| = \left| { - 4{\omega ^2}} \right|\)

3ω = +4ω²

\(\Rightarrow \omega = \left( {\frac{3}{4}} \right)rad/s\)

So, time period,

\(T = \frac{{2\pi }}{\omega }\)

\(\Rightarrow T = \frac{{2\pi }}{3} \times 4 = \frac{{8\pi }}{3}s\)

Therefore, its periodic time (in seconds) is \(\frac{{8\pi }}{3}\) s.