Frequency, Period, Displacement of Springs: Solving Problems

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In summary, the given particle has a displacement formula of x=3.0\cos(5\pi t+\pi), with x in meters and t in seconds. The frequency, F, and the period, T, of motion are respectively 2.5 and 0.4 seconds. The greatest distance the particle travels from equilibrium is 3.0 meters, and at time t=0.5 seconds, the particle is located at 3.0 meters. The velocity at t=0.5 seconds can be calculated by taking the derivative of the displacement formula, which results in a velocity of 0. Thank you.
  • #1
bard
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A particle has displacement given by[tex]x=3.0\cos(5\pi t+\pi)[/tex] 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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  • #2
Apply the formula for velocity i.e dy/dx
 
  • #3
I have no idea what you are talking about? are my other answers correct?
 
  • #4
d)no idea

Others are correct i thought you need to calculate the velocity
dont u
 
  • #5
yes i need to calculuate velocity, but i don't know how to get it.
 
  • #6
Look you got the displacement, Pls tell yourself what is the relation between displacement and velocity
 
  • #7
ok so i have to take the derivative

[tex]x=3.0\cos(5\pi t+\pi)[/tex]

so the derivative of that would be

[tex]3.0*-\sin(5\pi(0.5)+\pi)*d/dx(5\pi(0.5)+\pi[/tex]
 
  • #8
Yes You got it right
 
  • #9
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
[tex]3.0*-\sin(5\pi(0.5)+\pi)*d/dx(5\pi(0.5)+\pi[/tex]

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

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
 

What is frequency and how is it related to springs?

Frequency is the number of oscillations or cycles a spring completes in one second. It is directly proportional to the stiffness of the spring and inversely proportional to its mass. This means that a stiffer spring or a lighter mass will result in a higher frequency.

What is the period of a spring?

The period of a spring is the time it takes for one complete oscillation or cycle. It is the inverse of frequency, meaning that the higher the frequency, the shorter the period and vice versa.

How is displacement of a spring calculated?

Displacement of a spring can be calculated using the equation x = A * sin(2πf * t), where x is displacement, A is amplitude, f is frequency, and t is time. This equation is based on the properties of simple harmonic motion.

How can I solve problems involving springs and their frequency, period, and displacement?

To solve problems involving these variables, you can use the equations x = A * sin(2πf * t) for displacement, T = 1/f for period, and f = 1/T for frequency. You will also need to know the values of other variables such as mass, spring constant, and amplitude.

What factors can affect the frequency and period of a spring?

The frequency and period of a spring can be affected by factors such as the stiffness of the spring, the mass attached to the spring, and the amplitude of the oscillations. Other external forces, such as dampening or external forces, can also affect the frequency and period of a spring.

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