Struggling with finding the mass req'd when velocity is doubled

In summary: Your approach is correct. You could also try using conservation of energy.So using the new VmaxVmax = Aω0.516 = 0.01ωω = 0.516/0.01ω = 51.6ω = √(k/m)51.6 = √(200/m)m = 200/51.62m = 0.075kg
  • #1
Harrison01
26
0

Homework Statement


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.


Homework 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



The Attempt at a Solution



This is where i struggle as I am having a blonde moment and just can't 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)
 
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  • #2
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
Your approach is correct. You could also try using conservation of energy.
 
  • #4
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
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
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
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
OK, I see.

I suppose only using numbers at the end reduces the chance for error througout the workings as well.
 
  • #9
Harrison01 said:

Homework Statement


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.


Homework 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



The Attempt at a Solution



This is where i struggle as I am having a blonde moment and just can't 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)

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?
 

1. How does doubling the velocity affect the required mass?

Doubling the velocity will result in a higher required mass. This is because the kinetic energy of an object is directly proportional to its velocity squared. Therefore, doubling the velocity will result in four times the kinetic energy, requiring a higher mass to achieve the same amount of force.

2. What is the formula for calculating the required mass when velocity is doubled?

The formula for calculating the required mass when velocity is doubled is: Mass 1 * (Velocity 1 / Velocity 2)^2 = Mass 2. This means that the initial mass is multiplied by the ratio of the initial velocity to the final velocity, squared.

3. Can the required mass be calculated if the velocity is tripled or quadrupled?

Yes, the required mass can be calculated if the velocity is tripled or quadrupled. The formula remains the same, but the ratio of velocities will change accordingly. For example, if the velocity is tripled, the ratio would be (Velocity 1 / Velocity 3)^2.

4. How does the type of force being applied affect the required mass?

The type of force being applied does not affect the required mass when velocity is doubled. The required mass is solely determined by the change in velocity and the initial mass.

5. Can the required mass be reduced by increasing the velocity even more?

No, the required mass cannot be reduced by increasing the velocity even more. The required mass will continue to increase as the velocity increases, according to the formula mentioned in question 2.

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