What frequency of vibration is required?

In summary, the equation for speed is f*w. The equation for acceleration is -2*pi*f*x. The maximum speed is 100*v*w. The time delay between the points of maximum acceleration and maximum speed is t*-127.32*v.
  • #1
imy786
322
0

Homework Statement



(a) acceleration amplitude of 100 ms−2 with a vibration amplitude of 5.0 cm.

(i) What frequency of vibration is required?
(ii) What is the maximum speed ?
(iii)What is the time delay between the points of maximum acceleration and
of maximum speed?

Homework Equations



speed= f * wavelength

f= 1/ time period

The Attempt at a Solution



(i)

a= -2*pi*f*x

100= -2*pi*f*0.05

400=-pi *f

400/ -3.141= f

f= - 127.32 Hz

(can frequency be minus)??

(ii)

speed = integral of accelration
a= -2*pi*f*x

speed= f * wavelength

(iii)

time delay
f= 1/ time period
t= 1/ - 127.32 Hz
t= 0.00785423
 
Last edited:
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  • #2
Seems to me like this is a problem in simple harmonic motion. The acceleration to be found is the acceleration of a point on the medium that the wave is traveling through. As a wave travels through a medium it causes the particles within that medium to oscillate, each one about their own equilibrium position.
Smple harmonic motion is the simplest form of oscillation. Its defining equation is:-

a= -2*pi*f*x

where
a = acceleration
f= frequency
x= displacement from equilibrium position (max value = amplitude)

Hope this gets you started.
 
  • #3
a= -2*pi*f*x

100= -2*pi*f*0.05

400=-pi *f

400/ -3.141= f

f= - 127.32 Hz

(can frequency be minus)??
 
  • #4
(ii)

speed = integral of accelration
a= -2*pi*f*x

speed= f * wavelength

f= - 127.32 Hz
 
  • #5
(iii)

time delay
f= 1/ time period
t= 1/ - 127.32 Hz
t= 0.00785423

is this correct (i, ii, iii)
 
  • #6
well is this correct?
 
  • #7
"a= -2*pi*f*x"

This is not an equation for acceleration. Check the units, they don't work out. Maybe you didn't copy it from your book correctly? Go look it up again.

well is this correct?

Please try to be patient about waiting for replies. :smile:
 
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  • #8
(a) acceleration amplitude of 100 ms−2 with a vibration amplitude of 5.0 cm.
(i) What frequency of vibration is required?

u gave me this equation at top of this page=

a= -2*pi*f*x

i have used it...doesnt work..??
 
  • #9
a= -2*pi*f*x

where
a = acceleration
f= frequency
x= displacement from equilibrium position (max value = amplitude)

Hope this gets you started.
--------------------------------------------------------

a= -2*pi*f*x
100=-2*3.141*0.05* f
f= 317 Hz...

is this correct...using the formula u gave
 
  • #10
imy786 said:
a= -2*pi*f*x

where
a = acceleration
f= frequency
x= displacement from equilibrium position (max value = amplitude)

Hope this gets you started.
--------------------------------------------------------

a= -2*pi*f*x
100=-2*3.141*0.05* f
f= 317 Hz...

is this correct...using the formula u gave

I'm not the one that gave you the formula. You started off with it yourself by the looks of things. But it doesn't matter, it is still wrong.
Look at it carefully. The units work out to those of velocity, not acceleration. Check it. So, what would the right equation be? It's very similar.
 
  • #11
sorry HAGE,

apeeling gave me the INCORRECT acceleration formula,

well if the formula is velocity then u differtiate to get accelration.

But i don't want to start off with INCORECT formula.

Can you please give me the correct formula to start with.
-------------------------------------------------------------

w= 2*pi*f

accelration= -W^2 x

100= -w^2 * 0.05
2000= -W^2
+- square root of 2000= w

w=2*pi*f

+- square root of 2000= 2*pi*f

+- square root of 2000/2*pi= f

---------------------------------------
 
  • #12
"accelration= -W^2 x"

Get rid of the negative sign, I don't understand why it's there. Do that, and you have the right equation.
 
  • #13
formula booklet states a minus sign.

Does that depend where u take the point of refrence?

so doing this kind of problem you leave minus sign. What is the signifcan of minus sign again.

-----------------------------------------------------------------

(ii) What is the maximum speed ?
a= v w
100= v w

would that be correct method

---------------------------------------
(iii)What is the time delay between the points of maximum acceleration and
of maximum speed?

what formula needed to work at time delay
 
  • #14
imy786 said:
formula booklet states a minus sign.
Does that depend where u take the point of refrence?

so doing this kind of problem you leave minus sign. What is the signifcan of minus sign again.

-----------------------------------------------------------------

(ii) What is the maximum speed ?
a= v w
100= v w

would that be correct method

---------------------------------------
(iii)What is the time delay between the points of maximum acceleration and
of maximum speed?

what formula needed to work at time delay

For the maximum acceleration I would think it would be positive. But whatever, go by what you think.
(b): Yeah, i guess that works. You could also use the equation for speed that you were using by accident for acceleration.
(iii): Have you tried to come up with it yourself?
 
  • #15
(iii)
x= Asin (wt)

v= Awcos (wt)
100/w= v

100= Aw^2 (wt)

A is a constant, what will this constant be?

would this be right equation for (iii)
 
  • #16
I don't understand what you're trying to show. What will what constant be? You can reason this out as well, without any math. So if you can get both to agree then you probably have it right.

Do watch out for the signs in this, you just have to be careful what case you are taking (like at max acceleration, or just at some point x that is not at max acceleration). Sorry I know that isn't very clear, but I think you're aware of it anyway.
 
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  • #17
well is was that the corect formual to work out time delay.

i quoted the formula...from formula booklet
 
  • #18
need help on this question??
 
  • #19
i)
w= 2 pi f

accerleation= A w^2
100 = 0.05 * w^2
w= 44.721

44.721= 2 pi f
f= 7.11 Hz

ii)
V= Aw
= 0.05* 44.721
=2.23 m/s

is this corect?
 
  • #20
urgent help required is this CORRECT?
 
  • #21
imy786 said:
i)
w= 2 pi f

accerleation= A w^2
100 = 0.05 * w^2
w= 44.721

44.721= 2 pi f
f= 7.11 Hz

ii)
V= Aw
= 0.05* 44.721
=2.23 m/s

is this corect?

I would say those are correct.
 
  • #22
(iii)
x= Asin (wt)

v= Awcos (wt)
100/w= v

100= Aw^2 (wt)


is this correct method
 
  • #23
imy786 said:
(iii)
x= Asin (wt)

v= Awcos (wt)
100/w= v

100= Aw^2 (wt)


is this correct method

You keep quoting this formula from your formula booklet. Can you explain what you are doing with it? How are you getting the time difference out of this? Work it out completely, so I can see what you're doing.

What you are looking for is the time difference between when the maximum velocity is reached and when the maximum acceleration is reached. So, you need to figure out the relationship between the two. Try drawing out the x(t), v(t) and a(t) functions to see the this. Once you see what the phase is between them (how they are shifted with respect to one another), you can figure out the time difference because you know the frequency (and therefore the period) of the motion.
 
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  • #24
x= Asin (wt)
v= Awcost sin (wt)
aceleration=-Aw^2 sin wt

phase differnce between them would be of 90 degrees (the differnce between sin and cos)

still need more help to do question (iii)

(iii)What is the time delay between the points of maximum acceleration and
of maximum speed?
 
  • #25
OK, so you know the phase difference. How much of a wavelength does 90 degrees represent? So, how much time has passed in 90 degrees (keep in mind you know the period, so you already know how much time it takes for one full wavelength). Figure that out and you have your answer.
 
  • #26
(iii)What is the time delay between the points of maximum acceleration and
of maximum speed?

time delay between max speed and acceleration is =

t= pi/2

is this correct
 
  • #27
please someone help me do this question URGENT...
 
  • #28
You know the actual period in seconds of the wave right? You found that a long time ago. So at pi/2 how many seconds have gone by?
 
  • #29
actuly i don't know the time it takes, iv looked back and i havnt calculated it.

do i use these 2 formulas to work out the time for each the velocity and accereation-

v= Awcost sin (wt)
aceleration=-Aw^2 sin wt
 
  • #30
What's the relationship between frequency and period? I know you've written that down somewhere before, go back to your very first post. So, if you know the frequency (which you have calculated as part of (i)), you can get the period.
 
  • #31
f=1/t
t= 1/f

= 1/ 7.11
t= 0.14 seconds...

is this the time delay...then?
 
  • #32
NO that's the period. Do you know what that means? How much of the period goes by in pi/2 of the wavelength? That's what you need to get.
 
  • #33
not sure...
well i know speed= frequncy * wavelength
 
  • #34
No. OK, the period is the time for one wavelength. You have that now. So if pi/2 (or 90 degrees) of the wavelength goes by, how much time is that? Basically how much time for 1/4 of the wavelength? You don't need velocity for this.
 
  • #35
f=1/t
t= 1/f

= 1/ 7.11
t= 0.14 seconds

0.14/4= 0.035 sec
 

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