Frequency, Period, Displacement of Springs: Solving Problems

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The discussion revolves around solving a particle's motion described by the equation x=3.0cos(5πt+π). The frequency (F) is calculated to be 2.5 Hz, and the period (T) is 0.4 seconds. The maximum displacement from equilibrium is 3.0 meters, and the particle's position at t=0 and t=0.5 seconds is also 3.0 meters. Participants express confusion about calculating velocity, emphasizing the need to take the derivative of the displacement function. The correct approach to find velocity involves applying the time derivative to the displacement equation.
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A particle has displacement given byx=3.0\cos(5\pi t+\pi) where x is in meters and t in seconds

A)What are the frequency ,F, and the period,T, of motion
B)What is the greatest distance the particle travels from equlibrium
c)Where is the particle at time t=0?t=0.5(s)
D)What is the velocity when t=0.5s

A)F=2.5, T=0.4
B)3.0M
c)3.0M
d)no idea

thank you
 
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Apply the formula for velocity i.e dy/dx
 
I have no idea what you are talking about? are my other answers correct?
 
d)no idea

Others are correct i thought you need to calculate the velocity
dont u
 
yes i need to calculuate velocity, but i don't know how to get it.
 
Look you got the displacement, Pls tell yourself what is the relation between displacement and velocity
 
ok so i have to take the derivative

x=3.0\cos(5\pi t+\pi)

so the derivative of that would be

3.0*-\sin(5\pi(0.5)+\pi)*d/dx(5\pi(0.5)+\pi
 
Yes You got it right
 
yes but i get v=0 when i do the derivative. is this correct?

thnx for ur help
 
  • #10
Sorry U didn't write correctly
3.0*-\sin(5\pi(0.5)+\pi)*d/dx(5\pi(0.5)+\pi

it should be
3.0*-\sin(5\pi(0.5)+\pi)*d/dt(5\pi(0.5)t+\pi

I want to correct myself
Apply the formula for velocity i.e dy/dx

the velocity is time derivative
i.e v = dx/dt.

i just wanted to point towards the calculus involved
 
  • #11
thnx a lot man
 
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