Simple Harmonic Motion Homework: Amplitude 4.0cm, Frequency 50Hz

[imy786]

Homework Statement

amplitude is kept at 4.0 cm and the vibration
frequency increased to 50 Hz. T
The particle passes through the origin at time t = 0.0 s, traveling in the +y-direction.

(i) At what time does the valve first pass through the position y = 3.0 cm?
(ii) What is the speed at this point?
(iii) What is the acceleration at this point?
(iv) At what positions will the valve have a speed of 6.0ms−1?

Homework Equations

w= 2*pi*f
acceleration= -w^2*x
v= rw

The Attempt at a Solution

w= 2*pi*f
= 2*50*pi= 100pi


acceleration= w^2*x
= 100pi*4= 400pi

dont know if this is correct?
also don't know how to get the time and postions part (i and iv)

 

[dimensionless]

What is the equation for the motion?

What do you call the time derivative of the equation of motion?

 

[dobry_den]

(i) At what time does the valve first pass through the position y = 3.0 cm?
(iv) At what positions will the valve have a speed of 6.0ms−1?

also don't know how to get the time and postions part (i and iv)

ad i) Do you know the equation for simple harmonic motion - i.e. the relation between position and time? Solving this part is just a matter of solving this equation for y = 3.

ad iv) The same story. Just use the equation* for velocity of simple harmonic movement and solve for v = 6 m/s to get the time and use the equation in i) to get the position.

*you might get this one by differentiating the equation in i)

acceleration= -w^2*x

acceleration= w^2*x
= 100pi*4= 400pi

dont know if this is correct?

don't forget the minus sign - do you know what its significance is?

 

[imy786]

i)

equation of motion
v2= u2+2as

w= 2pi f
w= 100 pi

equation for SHM:x= asin wt
3= 4 sin 100 pi t
0.75= sin 100pi t
inverse 0.75= 100pi t
48.590...= 100 pi t
t= 1.527 secs

 

[imy786]

- indicates acceleartion towards the starting position

 

[imy786]

ii)
v= rw
= 3pi

 

[imy786]

iv)

a= vw
= 600pi

is questions i- iv correct?

 

[Mentz114]

i) is correct.
ii) is wrong. To get a velocity differentiate the x=asinwt wrt t.

 

[imy786]

ii)v= awcos wt

a= 3
w= 100pi
t=0
using this values would this be ok ?
and what about
iii)
iv)

 

[imy786]

ii)

coz0= 1
v= 300 pi

is this correct?

 

[nithin]

Ok for the first one , you need to form a equation and then solve it . It is given that the y - position is 0 at t=0. w=100pi and A=0.04m (always work with s.i units) . Using these info we can form the equation , y=A Sin(wt) , in this case it would be 0.03=0.04Sin((100pi)(t)) , solving for this will result in the answer!

 

[nithin]

Sorry about the double post

 

[nithin]

For part 2 , once again we need to form a equation, this time , we need to form the equation for velocity. Which would be v=-wACos(wt) (it is cos because at origin , y=0) From i) you have your t which you can sustitute into the equation and solve it!

 

[nithin]

Ok for part 3) , you need to form another equation. The equation would be a=-W^2ASin(wt). Using the ans from part 1) , you can sole the equation and get your ans!

 

[nithin]

just for you info imy786 , v=Aw and a= Aw^2 give you the maximum velocity and maximum acceleration.

 

[imy786]

nithin...

v=-wACos(wt

i don't think it has a minus sign.

book quotes

v= Awcos wt

 

[imy786]

also

a= Aw^2 sin (wt)
v= Awcos wt

you wrote these equations -

v=-wACos(wt
a=-W^2ASin(wt

 

[imy786]

URGENT NEED help...

 

[nithin]

imy786 it depends on the situation of the question . Sketch a simple y vs t and v vs t graph side byside ,You will then see whether it is - or +

 

[nithin]

I use cos instead of sin in my equations as my phase angle is 0 as it starts from y=0 at t=0. If u wish u can also use y= A Sin(wt + phi) then v= -AWCos(wt + phi) .Ooops my bad . Sorry .Like i said just draw a graph and you will see what to do.

 

[nithin]

If you are confused on what sign to use , just use calculus and see what function you get when u differentiate your y = smth smth function. My bad , i am just learning calculus.

 

[imy786]

i)

w= 2pi f
w= 100 pi

equation for SHM:x= asin wt
3= 4 sin 100 pi t
0.75= sin 100pi t
inverse 0.75= 100pi t
48.590...= 100 pi t
t= 1.527 secs

i was told this is correct...(but it wouldn't be if i was suppose to use RADIAN mode on calculator)...

then it would be-

3= 4 sin 100 pi t
0.75= sin 100pi t
inverse sin 0.75 = 100 pi t
0.848062= 100 pi t
t= 2.699*10 ^-3 seconds...

is this correct?

?

 

[imy786]

please someone help me do this question URGENT...

 

[denverdoc]

Radians are a natural if unintuitive measurement. 360 degrees are equal to 2*pi radians, about 6.28. If it helps think of a radian as 360/(2Pi)= 57.3 degrees. Make sure you know how to make your calcultor behave in both modes. I was intrigued by the fact that a Colo School of Mines physics test actualy prompted the test taker to convert to radians and back again. Try to get comfortable with both units so that you don't need someone to tell you to do so. So probably an unhelpful post.

Just a PS: nothing re schoolwork is ever urgent, and nothing approaching understanding ever answers such a call. Not to ridicule or dismiss any anxiety... Been there done that, and i can say in retrospect that any wrong or late physics answer has had no impact on my life. Just getting your head clear of this stress may help.

 

[imy786]

i)

w= 2pi f
w= 100 pi

equation for SHM:x= asin wt
3= 4 sin 100 pi t
0.75= sin 100pi t
inverse 0.75= 100pi t
48.590...= 100 pi t
t= 1.527 secs

i was told this is correct...(but it wouldn't be if i was suppose to use RADIAN mode on calculator)...

then it would be-

3= 4 sin 100 pi t
0.75= sin 100pi t
inverse sin 0.75 = 100 pi t
0.848062= 100 pi t
t= 2.699*10 ^-3 seconds...

is this correct doesn't the time look tooo small?

??

 

[hage567]

Well I can’t see how this is correct, even if it is in degrees. You are off by a factor of 10, double check it. Watch out for your rounding. Doesn’t matter though, it’s wrong anyway. Radians is the way to do this.

Think about this:

Look at the period of the wave, it’s 0.02s (remember, you can find this from the frequency). Is it reasonable for the first value of time passed at y=3 to be 1.527s if it only takes 0.02s for the entire wavelength to go by? I will tell you that it isn’t. Now, do you still think your answer in radians is too small? Even if you had found the “right” value for t (in degrees), does it make sense?

Like I said, use radians. Who told you that answer was correct? I don’t believe that it is. So I would like to know the source.

You must work on developing your physical understanding of what is happening in these problems, it will make it much easier for you to judge if your answer is a reasonable one.

 

[imy786]

Mentz114 it is correct??

POST 8......

 

[hage567]

When I do it (in degrees), I get:

3=4sin100pi*t
48.59=100*pi*t

So then

t = 48.59/(100*3.14159) = 0.1547s

not 1.527s. I don't see how you got this.

Go with your answer in radians, the 0.0027s. It’s the right one.

 

[imy786]

hage thank you so much for your help in the last few weeks...

thanks millions...