two collinear harmonic oscillations are X1= 8 sin( 100pi *t) and X2= 12 sin(96 pi *t) are superposed. calculate the values of time when the amplitude of the resultant oscillation will be maximum and minimum
To calculate the values of time when the amplitude of the resultant oscillation is maximum and minimum, we need to find the resultant oscillation by superposing the two given oscillations.
The equation for the resultant oscillation can be obtained by summing the two individual oscillations:
X = X1 + X2
Substituting the given values:
X = 8sin(100πt) + 12sin(96πt)
To find the time values when the amplitude of the resultant oscillation is maximum and minimum, we need to find the moments when the derivative of X with respect to time, dX/dt, is equal to zero.
Let's calculate dX/dt:
dX/dt = 8*(100π)cos(100πt) + 12*(96π)cos(96πt)
Now, set dX/dt to zero and solve for t:
8*(100π)cos(100πt) + 12*(96π)cos(96πt) = 0
Dividing through by 4π:
2*200cos(100πt) + 3*192cos(96πt) = 0
Simplifying further:
400cos(100πt) + 576cos(96πt) = 0
To solve this equation, we can use the sum-to-product formula for cosine:
cos(A) + cos(B) = 2*cos((A+B)/2)*cos((A-B)/2)
Using this formula, we can rewrite the equation as:
2*cos((100πt + 96πt)/2)*cos((100πt - 96πt)/2) = 0
Now, we have two cases to consider:
1. When cos((100πt + 96πt)/2) = 0:
(100πt + 96πt)/2 = (2n + 1)*(π/2) for n = 0, ±1, ±2, ...
Simplifying:
196πt = (2n + 1)*(π/2)
t = (2n + 1)*(π/2*196π)
Simplifying further:
t = (2n + 1)/(392)
2. When cos((100πt - 96πt)/2) = 0:
(100πt - 96πt)/2 = (2n + 1)*(π/2) for n = 0, ±1, ±2, ...
Simplifying:
2πt = (2n + 1)*(π/2)
t = (2n + 1)*(π/4)
So, the time values when the amplitude of the resultant oscillation is maximum and minimum are given by the equations:
1. Maximum amplitude: t = (2n + 1)/(392)
2. Minimum amplitude: t = (2n + 1)*(π/4)
where n is an integer. Plug in the values of n to find the specific time instances.