Asked by veena
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:
Answers
Answered by
MathMate
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.
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.
Answered by
Jeff
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.
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.
Answered by
7mbola
lol
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