A body oscillates with simple harmonic motion along the x axis. Its displacement varies with time according to the equation x =5sin πt + π 3( ). The velocity of the body at t = 1.0 s is (-8 m sec )
What is the question?
To find the velocity of the body at t = 1.0 s, we can differentiate the given displacement equation with respect to time.
Given:
x = 5sin(πt + π/3)
Differentiating both sides with respect to time, we get:
v = dx/dt = d/dt (5sin(πt + π/3))
To differentiate the sine function, we use the chain rule. The derivative of sin(u) with respect to u is cos(u), and the derivative of u with respect to t is du/dt.
So, applying the chain rule, we have:
v = d/dt (5sin(πt + π/3))
= 5 * cos(πt + π/3) * d/dt (πt + π/3)
Since d/dt (πt + π/3) is π (the derivative of t is 1), we have:
v = 5 * cos(πt + π/3) * π
Now, we can substitute t = 1.0 s into the equation to find the velocity at that time:
v = 5 * cos(π(1.0) + π/3) * π
= 5 * cos(π + π/3) * π
= 5 * cos(4π/3) * π
= 5 * (-1/2) * π
= -5/2 * π
Therefore, the velocity of the body at t = 1.0 s is -5/2 * π m/s or approximately -7.85 m/s.