A body is executing s.h.m. of amplitude 1m .Its velocity while passing through the mean position is 10m/s.Find its frequency.
v(max) = v =A•ω=A•2πf
f = 2Aπ/v
Answer for this question is what????
To find the frequency of a body executing Simple Harmonic Motion (SHM), we need to use the formula:
frequency = (1 / time period)
In the given question, we are provided with the amplitude of the motion (A = 1m) and the velocity when the body passes through the mean position (v = 10m/s).
To find the frequency, we need to determine the time period of the motion first. The time period is the time taken for the body to complete one full oscillation. When the body is at the extreme position (maximum displacement), it momentarily stops before changing its direction. At this point, the velocity is zero. The velocity is maximum when the body passes through the mean position.
Using this information, we can calculate the time period by equating the maximum velocity (v) with the product of the angular frequency (ω) and the amplitude (A):
v = ωA,
where ω = 2πf (angular frequency) and f (frequency) = 1 / T (time period).
Let's solve for the time period (T):
v = ωA
10 = 2πf * 1
10 = 2πf
f = 10 / (2π)
f ≈ 1.59 Hz
Therefore, the frequency of the body executing SHM is approximately 1.59 Hz.