A particle is in simple harmonic motion along the x-axis. The amplitude of the oscillation is A and its period is T. If at t = 0 the particle is at x = +A, then at t = 1.25T, the particle i
shaklo kamin inti 3ndak assignment physics lmao
Answer: At t=1.25T , the particle is at x=0 and is travelling toward x = -A
To determine the position of the particle at t = 1.25T, we can use the equation for simple harmonic motion along the x-axis:
x = A * cos(ωt + φ)
Where:
x = position of the particle at time t
A = amplitude of the oscillation
t = time
ω = angular frequency (2π/T)
φ = phase constant
Given that at t = 0, the particle is at x = +A, we can substitute these values into the equation:
A * cos(0 + φ) = A
Since cosine of 0 is equal to 1, we get:
A * cos(φ) = A
Dividing both sides by A, we have:
cos(φ) = 1
This implies that the phase constant φ is 0, since the cosine of 0 is 1.
Now, let's evaluate the position of the particle at t = 1.25T:
x = A * cos(ωt + φ)
x = A * cos(ω * 1.25T)
Since ω = 2π/T, we can substitute this value into the equation:
x = A * cos((2π/T) * 1.25T)
x = A * cos(2π * 1.25)
x = A * cos(2.5π)
The cosine of 2.5π is -1, so we find:
x = A * (-1)
x = -A
Therefore, at t = 1.25T, the particle is at x = -A.
we want max at t = 0 so use cosine
x =A cos (2 pi t/T)
at t = 1.25 T, same as t = 0.25 T
x = A cos(2 pi * .25) = A cos pi/2 = A * 0 = 0