# Homework Help: Struggling with finding the mass req'd when velocity is doubled

1. Oct 7, 2012

### Harrison01

1. The problem statement, all variables and given/known data
Hey peeps, I'm really struggling with this one and i'm looking for a point in the right direction.

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,

(iv) the mass required to produce double the maximum velocity calculated in (ii) using the same spring and initial deflection.

2. Relevant equations

Am i right in assuming that i need to use a transposition of the following formula to find mass,
f=1/2pi√ k/m

3. The attempt at a solution

This is where i struggle as im having a blonde moment and just cant seem to start this one.
so far i have,
frequency of 4.109
time of 0.243
velocity of 0.25819
accelecration of 6.666
New velocity of 0.51638 (2x.25819)

2. Feb 4, 2013

### Carlo2986

I'm no expert so please don't take my answer as law, but I worked it out going back from the original Vmax of 0.258ms-1

You need to find the mass that will give you double your original Vmax which will be 0.516ms-1

I used equations Vmax=Aω and ω=√(k/m)

Im sure someone could correct me if I'm wrong?

3. Feb 4, 2013

### voko

Your approach is correct. You could also try using conservation of energy.

4. Feb 4, 2013

### Carlo2986

So using the new Vmax

Vmax = Aω
0.516 = 0.01ω
ω = 0.516/0.01
ω = 51.6

ω = √(k/m)
51.6 = √(200/m)
m = 200/51.62
m = 0.075kg

Does that look correct?

5. Feb 4, 2013

### voko

That is correct, but you do not really have to get the numeric value of Vmax and ω. Vmax = Aω, and 2Vmax = Aω', so 2√(k/m) = √(k/m'), yielding m' = m/4 = 0.3/4 = 0.075.

6. Feb 4, 2013

### Carlo2986

Unfortunately i'm not quite as gifted so I have to go the long way round just so it makes sense. When you use " ' " what does that mean?

7. Feb 4, 2013

### voko

I used m' to denote the mass that has twice the max velocity, and ω' means the angular frequency corresponding to twice the max velocity.

Speaking of giftedness, that's not it, it just takes a little practice. You should always try to solve problems symbolically throughout, and plug the numbers only at the very end.

8. Feb 4, 2013

### Carlo2986

OK, I see.

I suppose only using numbers at the end reduces the chance for error througout the workings as well.

9. Feb 4, 2013

### PeterO

Seems to me that you are doing the same amount of work on the spring, when displacing it form the equilibrium position, in each case. That should store the same amount of energy in the spring, which will be converted to maximum Kinetic energy when the mass passes through the equilibrium position.

That means the mass will have the same amount of kinetic energy each time, so mv2 is the same both times.

If the speed is to be doubled, the contribution of the v2 is up by a factor of 4, so it would seem the mass has to be reduced by a factor of 4 to compensate.

Perhaps it is not that simple?