A body oscillates with simple harmonic motion along the x-axis. Its displacement varies with time according to the equation x=5 sin (PIE + PIE/3). The velocity (in m/s) of the body at t=1 second is:
The original problem asked for the velocity at t= 1s of the position x(t)= 5sin(pt + p/3).
Take the derivative of the position function dx(t) = v(t) = 5pCos(p + p/3)
v(t) = -5*p*Cos(p + p/3)
v(t) = -7.9
The negative sign is part of the equation for the angular frequency of the velocity of the body.
There's probably a typo, you probably meant to post
x(t)=5 sin (π + πt/3)
The velocity is the derivative of x with respect to time, giving:
velocity=x'(t)=5(π/3)cos(π+πt/3)
Substitute t=1 in x'(t) to get the velocity at t=1.
lol
Well, let's break it down. The equation given is x=5 sin (π + π/3).
Now, the velocity of an object undergoing simple harmonic motion is given by the derivative of its displacement with respect to time.
So, let's differentiate x with respect to t:
dx/dt = 5 cos (π + π/3)
Now, let's simplify this:
cos (π + π/3) = cos π cos (π/3) - sin π sin (π/3)
Since cos π = -1 and sin π = 0, we can simplify further:
cos (π + π/3) = -1 cos (π/3)
Finally, let's evaluate this:
cos (π/3) = 1/2
Now, substituting this back into our differentiation equation:
dx/dt = 5 * (1/2)
dx/dt = 5/2
So, the velocity of the body at t=1 second is 5/2 m/s.
Now, that answer is as serious as a circus, isn't it?
To find the velocity of the body at t = 1 second, we need to differentiate the displacement equation with respect to time.
Given the displacement equation x = 5 sin(πt + π/3), we can differentiate it to find the velocity equation.
Velocity (v) is the derivative of displacement (x) with respect to time (t), which can be represented as:
v = dx/dt
Differentiating the equation x = 5 sin(πt + π/3) with respect to t:
v = d/dt (5 sin(πt + π/3))
To differentiate a sine function with respect to time, we need to apply the chain rule. The derivative of sin(u) with respect to u is cos(u), and we multiply it by the derivative of the argument inside the sine function.
Applying the chain rule:
v = 5 * cos(πt + π/3) * d(πt + π/3)/dt
The derivative of πt + π/3 with respect to t is simply π since the derivative of t is 1.
v = 5 * cos(πt + π/3) * π
Now we can substitute t = 1 second into the velocity equation to find the velocity of the body at t = 1 second:
v = 5 * cos(π(1) + π/3) * π
Simplifying the equation:
v = 5 * cos(π + π/3) * π
Using the trigonometric identity cos(π + π/3) = -cos(π/3):
v = 5 * (-cos(π/3)) * π
Since cos(π/3) = 1/2, we can substitute its value:
v = 5 * (-1/2) * π
Simplifying the equation:
v = -5π/2
Therefore, the velocity of the body at t = 1 second is -5π/2 m/s.