A particle executing simple harmonic motion has a maximum speed of 40 ~ms ⁻¹ and maximum acceleration of 60 ~ms ⁻² . The period of oscillation is

Step-by-step solution

Given, maximum speed of SHM is 40 ~ms ⁻¹ . v_ =a =40 ~ms ⁻¹ ...(i) Maximum acceleration of SHM is 60 ~ms ⁻² . As, a_ =a ^2=60 ~ms ⁻²...(ii) Dividing Eq. (ii) by Eq. (i), we get array rlrl & & ^2 & = 60 40 & & & = 3 2 ~s ⁻¹ We know, & & =2 v & 3 2 & =2 1 T 1 T = 3 4 & & T & = 4 3 ~s array Hence, time period of the oscillation is 4 3 ~s .