Thread: Mass required to produce double velocity

1. Mass required to produce double velocity

A mass of 0.3 kg is suspended from a spring of stiffness 200 N m–1. If
the mass is displaced by 10 mm from its equilibrium position and
released, for the resulting vibration, calculate:
(a) (i) the frequency of vibration. For this I got 4.1 Hz
(ii) the maximum velocity of the mass during the vibration. I got 0.2582ms^-1 or 0.26ms^-1 (whichever you prefer)
(iii) the maximum acceleration of the mass during the vibration. I got 6.667ms^-2 or 6.7ms^-2 (whichever you prefer)
(iv) the mass required to produce double the maximum velocity. I don't know how to solve this and would like a little help please.....
calculated in (ii) using the same spring and initial deflection.
(b) Plot a graph of acceleration against displacement (x) (for values of x
from x = –10 mm to x = +10 mm). And like wise not sure how to go about this either. Could someone confirm my other answers are correct also please for peace of mind......

2. Re: Mass required to produce double velocity

I don't agree with your first answers. If we let w = sqrt(k/m), then frequency is calculated from 2 pi w. The next two answers are correct, which is interesting because if you got the frequency wrong I would expect that these others would be wrong as well.

Given that max velocity is V= Aw, and w is proportional to the inverse square root of m, if you decease m you get a larger value of w and hence a larger value of max velocity.

The graph question is a bit of a trick question. Given a displacement function of x(t)=A sin(wt), take its derivative to get the velocity equation v(t)=Aw cos(wt) and the second derivative gives a(t)=-Aw^2sin(wt). Can you see now how acceleration elates to displacement?

3. Re: Mass required to produce double velocity

so you think the frequency should be 2pi x 4.1 = 25.76 Hz? Completely lost with iv and b though my heads spinning!!...

4. Re: Mass required to produce double velocity

Show us your work for calculating w=sqrt(k/m); that's where the problem lies.

5. Re: Mass required to produce double velocity

w=sqrt(200/0.3) = 25.82 rads^-1 for the frequency I used 1/2pi sqrt(200/0.3)=4.1Hz

6. Re: Mass required to produce double velocity

Yes, I see - my mistake. Freq = 25.82 rad/sec divided by 2 pi cycles/radian = 4.11 Hz. You are correct -sorry for stating otherwise.

7. Re: Mass required to produce double velocity

That's ok and good to know my other answers are correct. Any idea how to solve the two which are not though? I haven't got a clue!!

8. Re: Mass required to produce double velocity

Consider the fact that the max velocity is equal to max displacement times w. Hence it is proportional to 1/sqrt(m). So if you decrease m the max velocity increases - do you see why that is? For example if you cut m in half the velocity increases by sqrt(2). From this you should be able to figure out how to change m to get the max velocity to double.

The graph I already described in an earlier post. Acceleration is w^2 times displacement, so you have a linear relationship between the two, regardless of the instantaneous displacement.

9. Re: Mass required to produce double velocity

Originally Posted by ebaines
Acceleration is w^2 times displacement ...
Sorry, should have written -(w^2).

calculate the veloclity req

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