The motion of a certain simple harmonic oscillator is described by: x(t) = (12 cm) sin [(1 s ^(-1)) t] What is the earliest time t > 0 or t = 0 at which the oscillator's velocity is zero?
To find the time when the oscillator's velocity is zero, we need to differentiate the given position function, x(t), with respect to time, t, to obtain the velocity function, v(t). Then, we can solve for the time when v(t) = 0.
Let's differentiate the position function, x(t), with respect to time:
x(t) = (12 cm) sin[(1 s^(-1))t]
To differentiate, we use the chain rule and the derivative of sin(x):
v(t) = d[x(t)]/dt = (12 cm) * cos[(1 s^(-1))t] * d[(1 s^(-1))t]/dt
Since d[(1 s^(-1))t]/dt is equal to (1 s^(-1)), we can simplify the expression to:
v(t) = (12 cm) * cos[(1 s^(-1))t]
Now, we want to find the time t at which v(t) = 0. Therefore, we set v(t) equal to zero and solve the equation:
0 = (12 cm) * cos[(1 s^(-1))t]
Now, we need to solve for t. To do this, we divide both sides of the equation by (12 cm):
0 = cos[(1 s^(-1))t]
To find the earliest time when cos[(1 s^(-1))t] is equal to zero, we need to find the smallest positive angle θ such that cos(θ) = 0.
The cosine function is equal to zero at θ = π/2, 3π/2, 5π/2, etc. Since the argument of the cosine function in our equation is (1 s^(-1))t, we solve for t as follows:
For θ = π/2:
(1 s^(-1))t = π/2
t = (π/2) / (1 s^(-1))
t = π/2 s
Therefore, the earliest time t > 0 or t = 0 at which the oscillator's velocity is zero is t = π/2 seconds.