A particle executes SHM with time period T and ampitude A.Find the maximum possible average velocity in time T/4
Maximum velocity is achieved twice in every period. The average V will depend upon which T/4 interval is selected. Since the interval is less than half the period, true maximum velocity may not be achieved during that short interval.
The maximum AVERAGE velocity in a T/4 interval will occur when it occurs in the middle of that interval.
Vav,max = Vmax*(Integralof)cos(2 pi t/T) dt/(T/4)
from t = -T/8 to T/8
Compute that integral for:
Vmax = 2 pi*(Amplitude)/T
8A/T
To find the maximum possible average velocity of a particle in Simple Harmonic Motion (SHM) within a time period of T/4, we need to understand the relationship between velocity and displacement in SHM.
In SHM, the velocity of a particle is influenced by its displacement from the mean position. As the particle moves away from the mean position, its velocity increases, reaches maximum at the extreme points, and then decreases as it returns to the mean position.
The time period (T) of SHM is the time taken for the particle to complete one full oscillation. The amplitude (A) represents the maximum displacement of the particle from its mean position.
To calculate the maximum possible average velocity within a time period of T/4, we need to know how the displacement varies with time during SHM. This can be derived from the equation of SHM:
x = A * cos(2πt / T)
where x is the displacement from the mean position at time t.
Differentiating the equation of displacement with respect to time, we get the expression for velocity:
v = dx/dt = (-2πA / T) * sin(2πt / T)
The maximum absolute velocity occurs at the extreme points of the SHM, where the sine function reaches its maximum value of 1.
Hence, the maximum velocity (vmax) can be calculated by substituting t = T/4 into the velocity equation:
vmax = (-2πA / T) * sin(2π(T/4) / T)
= (-2πA / T) * sin(π / 2)
= (-2πA / T)
Since average velocity is defined as the displacement divided by the time taken, the maximum possible average velocity within a time period of T/4 can be calculated as:
average velocity = vmax / (T/4)
= (-2πA / T) / (T/4)
= -8Aπ / T^2
Therefore, the maximum possible average velocity in time T/4 for a particle in SHM is equal to -8Aπ / T^2.