Question

The period of a simple harmonic oscillator is 2 sec. The ratio of it’s

maximum velocity & maximum acceleration is​

Answer 1

Answer:

The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is 10s−1. At, t=0 the displacement is 5m.

Answer 2

Given :

Time period of oscillation (T) = 2 sec

To Find :

The ratio of maximum velocity to maximum acceleration

Solution :

Angular Frequency (ω) = $\frac{2\pi }{T}$

                                       = $\frac{2\pi }{2}$

                                       = π rad/sec

Let 'A' be the amplitude of oscillation.

Maximum Velocity ($V_{max}$) = ωA

                                           = πA

Maximum acceleration ($a_m_a_x$) = ω²A

                                                   =  π²A

Ratio of $V_{max}$ to $a_m_a_x$ = $\frac{\pi A}{\pi ^2A}$

                                   = $\frac{1}{\pi }$

Therefore, the ratio of maximum velocity to maximum acceleration is found to be 1 : π