Simple harmonic motion and time

[nemzy]

A particle moving along the x-axis in simple harmonic motion starts from its equilibrium position, the origin, at t = 0 and moves to the right. The amplitude of its motion is 1.70 cm and the frequency is 2.00 Hz.

Determine the total distance traveled between t = 0 and t = 0.75 s.

how would you solve this?

well i can easily solve for T, w, A, etc..

but how can i find the total distance traveled between those 2 points? is there an easy formula i am overlooking?

i hate oscillations

 

[dextercioby]

HINT:
It's the difference between two values of "x".

Daniel.

 

[nemzy]

i am still clueless?

what do u mean the diff bw the two values of x?

how can i find the value of x at a given time??

all i know is the amplitude...but that isn't a value of x is it?

 

[dextercioby]

You know the amplitude,the frequency,u can compute the initial phase...
You know everything...

Daniel.

 

[HallsofIvy]

What is the period of the motion? Knowing that will tell you whether the motion was all in one direction, in which case the problem is easy, or if you need to take into account one or more changes of direction.

 

[nemzy]

do i use this formula to solve for x as a function of time?
x(t)= Acos(wt+&)

where w=angular frequency and &= initial phase?


in this case the t=the time right, not the period?

and...how do i know what & is?

when i have pi in cos or sin, do i use the radian or degree mode in my calc?

if i use radian or degree diff answers come out

 

[Jack_Legacy]

I'm not an expert and am currently studying this now so if it don't make no sense disreguard it okay: so we have f=2 T=0.75 A=1.7

so we could use the equation (x=distance)
x = sin A (2π × f × t) so: sin 0.17 (2π × 2 × 0.75) = 1.59m

SHM is always done using radians