A particle executing simple harmonic motion along a straight line has a distance of 24 cm between its two extreme positions. It crosses its mean position 60 times per minute. If the maximum velocity of the particle is v₀ cm/s , then the value of v₀ is _______.

Step-by-step solution

The distance between the extreme positions is twice the amplitude ( 2A ). 2A = 24 cm A = 12 cm In one complete oscillation, the particle crosses its mean position twice. Therefore, 60 crossings per minute corresponds to 30 oscillations per minute. The frequency f of the motion is: f = 30 60 = 0.5 Hz The angular frequency is: = 2 f = 2 0.5 = rad/s The maximum velocity of the particle is given by: v_ = A = 12 = 12 cm/s Comparing this with v₀ cm/s , we get v₀ = 12 . Answer: 12