# Frequency of vibration

#### DR1

1. Homework Statement

a mass attached to the lower end of a vertical spring causes the spring to extend by 25mm to its equilibrium position. The mass is then displaced a further 20mm and released. a trace of the vibration and time mesurements are taken. From these mesurements it can be seen that the displacement from the equilibrium position is 19.2mm when the time is 0.05s

A) calculate the expected frequency of vibration

B) calculate the maximum acceleration of the mass

C) calculate the maximum velocity of the mass

2. Homework Equations

i belive the frequency formule to be f=1/T Hz, or the displacement formule x=Asin(wt+o) both of which im sure are part of the solution but i can not see how to use them with the info given

3. The Attempt at a Solution

i am having an problem knowing were to start with this as i belive the key is finding out the weight of the Mass any help or pointers would be greatfully recived

many thanks

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#### kuruman

Homework Helper
Gold Member
You need to plug in the formula that you have for x(t) to find ω. If it is frequency f that you are looking for, them f = 2π/ω. Begin by identifying the symbols in the equation

x(t) = A sin(ωt+δ)

What is A, what is δ? Use the fact that at t = 0, the mass is at 20 mm from the equilibrium position and is not moving (instantaneously).

#### DR1

You need to plug in the formula that you have for x(t) to find ω. If it is frequency f that you are looking for, them f = 2π/ω. Begin by identifying the symbols in the equation

x(t) = A sin(ωt+δ)

What is A, what is δ? Use the fact that at t = 0, the mass is at 20 mm from the equilibrium position and is not moving (instantaneously).
ok so A is amplitude which is the maximum displacement = 0.02m

so putting all the info in gets 0.020=0.020sin(w0+angle)
therefore 0.02/0.02=sinw which is 0.017 this does not look right to me i am obviously missing something please help

#### kuruman

Homework Helper
Gold Member
ok so A is amplitude which is the maximum displacement = 0.02m

so putting all the info in gets 0.020=0.020sin(w0+angle)
therefore 0.02/0.02=sinw which is 0.017 this does not look right to me i am obviously missing something please help
No. Try again. What do you get if you solve 0.02 = 0.02 sin(ω*0+δ)? Any number multiplied by zero is zero, so there is no ω in the expression.

#### DR1

No. Try again. What do you get if you solve 0.02 = 0.02 sin(ω*0+δ)? Any number multiplied by zero is zero, so there is no ω in the expression.
ok so i am still mot following im afraid i am using the figures as
displacement (x) = 0.020
time (t)=0
Amplitude (A)=0.020
initial Angle (O)= 0
are these the correct figures to use or am i best using the figures once released ie

x=0.0192
t=0.05
A=0.02
O=0

i am trying to reaarange the formule for w but failing miserably as it is giving me silly answers

im sorry if i am missing the obvious

#### kuruman

Homework Helper
Gold Member
0.02 sin(ω*0+δ) = 0.02
sin(0+δ) = 0.02/0.02
sin(δ) = 1
δ = ?

#### DR1

0.02 sin(ω*0+δ) = 0.02
sin(0+δ) = 0.02/0.02
sin(δ) = 1
δ = ?
δ=90
if i were to then put this with the figures gained once released
0.0192=0.02sin(w*0.05+90) how do i get w on its own? do i need to mutiply out the barckets

#### kuruman

Homework Helper
Gold Member
ω*0.05+90=arcsin(0.0192/0.02)

#### DR1

ω*0.05+90=arcsin(0.0192/0.02)
ok so reaaranging for w i am getting

(arcsin(0.0192/0.02)/0.05)-90=1384.796 this sounds to big to me were am i going wrong

#### DR1

ω*0.05+90=arcsin(0.0192/0.02)
alternatively if i do ( Arcsin(0.0192/0.02))/90.05 = 0.82rads-1 this sounds better? could you confirm either way

#### kuruman

Homework Helper
Gold Member
ok so reaaranging for w i am getting

(arcsin(0.0192/0.02)/0.05)-90=1384.796 this sounds to big to me were am i going wrong
You put in the numbers incorrectly. Also, depending on the settings of your calculator, the arcsine may return the angle in degrees or radians. It's up to you to figure this out - I don't have your calculator in front of me.

alternatively if i do ( Arcsin(0.0192/0.02))/90.05 = 0.82rads-1 this sounds better?
Smaller but not necessarily better. How did that 90.05 get in the denominator? It seems you need to improve your algebra skills.

#### DR1

I am really struggling with this, my maths skills have never been particularly great as I am sure I am demonstrating!

ω*0.05+90=arcsin(0.0192/0.02)

You stated the above, to get w on its own I cannot work out whether I group up 0.05+90 and then put it in the denominator, or if I move both individually, and therefore put 0.05 in the denominator and minus 90 from the overall result?

I no longer have the book for my calculator so will try and find a manual online re the rads/degrees issue

Thanks for your assistance, it is appreciated, I know you must be getting frustrated with me!

#### kuruman

Homework Helper
Gold Member
If you are calculating the arcsine in radians, convert the 90 to π/2. Then move 90 (or π/2) over to other side and change its sign. Now divide everything on the right side and on the left side by 0.05. What do you get?

#### DR1

Am i right in say that my calculator is working out the arcsine in degrees as the result I get is 73.74?

#### kuruman

Homework Helper
Gold Member
Am i right in say that my calculator is working out the arcsine in degrees as the result I get is 73.74?
Yes you are right in that regard. Now proceed with the calculation of ω as I indicated.

#### DR1

(73.74-90)/0.05 = -325.2

alternatively if I do the rads option (having now found a guide for my calculator!) i get

(1.287002218-(π/2))/0.05 = 16.38

am i right in concluding that the first answer is in degrees and the second one is in rads so its the second one that I want?

#### kuruman

Homework Helper
Gold Member
Correct, it is rads that you want, but your numbers are wrong. What do you get when you subtract π/2 from 1.287?

#### DR1

-0.2838?

then dividing by 0.05 gives -5.676 as my answer?

#### DR1

and then dividing that by 2π gives me the answer in hertz?

#### kuruman

Homework Helper
Gold Member
Right. However, if you solve for the frequency, you get a negative number and that's not cool. Note that after one period T has gone by, the angle returns to the exact same value of -0.2838. This means that, for any time t, we can write
ω(t - T) = -0.2838.
Take out the parentheses and you get
ωt - ωT = -0.2838.
Now ωT = 2π. Put that in the equation, move it over to the other side and change its sign. Then solve for ω. You should get a positive value.

I know this sounds complicated. There is a simpler way to write x(t), but you should stick with this description until is it is finished. I don't want you to be unnecessarily confused.

Good luck. I have to sign off for now.

**** Edit ****
This posting refers to DR1's posting #18.

Last edited:

#### DR1

Ok thank you for that, in that case my answer is that w = 119.99 rads.

And in hertz I make the frequency 19.097?