How Do You Model the Motion of a Mass on a Spring?

In summary, the mass is displaced 19.2 mm as a result of the vertical spring being extended by 25 mm and then released. The displacement, velocity, and acceleration are all determined by the expressions y = A sin(ωt + Φ), v = Aw cos(wt + 0), and a = -Aw2 sin(wt + 0), respectively.
  • #1
Kev1n
40
0
1. A mass attached to the lower end of a vertical spring causes the spring to extend by 25 mm to its equilibrium position. The mass is then displaced a further 20 mm and released. A trace of the vibration and time measurements are taken. From these measurements it can be seen that the displacement from the equilibrium position is 19.2 mm when the time is 0.05 s.)
(a) Write the expression for the displacement of the mass as a function of time.
(b) Write the expression for the velocity of the mass as a function of time.
(c) Write the expression for the acceleration of the mass as a function of time.




2. a. Y= ASinWT, b. v= AwCoswT, c. a= -Aw^2Sin wT



3. I have calculated x = 0.0192, t = 0.05, w = 19.8, A = 0.02

Attempt

a. Y = 20Sin19.8t
b. v = 396Cos19.8t
c. a= -7840.8Sin19.8t

Any advice on where I am or what I am doing wrong appreciated, thanks
 
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  • #2
Since the mass is released at its maximum displacement from equilibrium, you'd be better off using cosine instead of sine to describe the displacement.

How did you compute ω?
 
  • #3
Doc Al said:
Since the mass is released at its maximum displacement from equilibrium, you'd be better off using cosine instead of sine to describe the displacement.

How did you compute ω?

W = sqrt K/M = sqrt 9810/25 = W^2 = 9810/25 = 19.8

I am actually quite lost as to what I have to achieve
 
  • #4
Kev1n said:
W = sqrt K/M = sqrt 9810/25 = W^2 = 9810/25 = 19.8
Good.
I am actually quite lost as to what I have to achieve
I presume the position and time were given so you can calculate the phase at t = 0. (I guess it isn't at maximum displacement at t = 0, as I had first thought. A bit strangely worded, I'd say.)

Use a more general form for the displacement; something like y = A sin(ωt + Φ).
 
  • #5
Doc Al said:
Good.

I presume the position and time were given so you can calculate the phase at t = 0. (I guess it isn't at maximum displacement at t = 0, as I had first thought. A bit strangely worded, I'd say.)

Use a more general form for the displacement; something like y = A sin(ωt + Φ).

Thanks, Do I insert values ie: A = 0.02xSin

So phase angle is 0
 
  • #6
Kev1n said:
Thanks, Do I insert values ie: A = 0.02xSin

So phase angle is 0

Then work out a figure for each epression e V = ****
 
  • #7
Kev1n said:
Thanks, Do I insert values ie: A = 0.02xSin

So phase angle is 0
No. You know A and ω. Insert the given values for y and t, then you can solve for the phase.
 
  • #8
Doc Al said:
No. You know A and ω. Insert the given values for y and t, then you can solve for the phase.

How does this look:

y = A sin(wt + b) - you are given x = 0.0192 when t = 0.05, A = 0.02 and you know w, so you can find a
d) y = A sin (wt + b)
A = 20 mm, w = 19.8 rad/s
Now y = 19.2 mm at t=0.05s
Substituting
19.2 = 20 Sin(19.8x0.05 + b)
Thus, Sin(0.99 + b) = 19.2/20
Sin(0.99 + b) = 0.96
0.99 + b = Sin-10.96 = 1.287 ( Take the angle in radians)
So b = 1.287 – 0.99 = 0.297
Or y = 20 Sin(19.8t + 0.297)

e) Similarly, v =396 Cos(19.8t + 0.297)

f) a = --7840.8 Sin(19.8t + 0.297)
 
  • #9
(a) Write the expression for the displacement of the mass as a function of time.

x = Asin (wt + 0)

(b) Write the expression for the velocity of the mass as a function of time.

V = dx/dt = Aw cos (wt + 0)


(c) Write the expression for the acceleration of the mass as a function of time.

a = dv/dt = -Aw2 sin (wt + 0)
 

FAQ: How Do You Model the Motion of a Mass on a Spring?

What are the different types of springs?

There are several different types of springs, including compression springs, extension springs, torsion springs, and constant force springs. Each type has its own unique shape and purpose.

How do springs work?

Springs work by storing potential energy when they are compressed or stretched. This energy is then released when the spring returns to its original shape, creating a force that can be used for a variety of purposes.

What are some common applications of springs?

Springs have many practical uses in everyday life, such as in car suspensions, door hinges, and pens. They are also used in more specialized applications like watches, mattresses, and medical devices.

How do I calculate the force of a spring?

The force of a spring can be calculated using Hooke's law, which states that the force applied to a spring is directly proportional to the distance it is stretched or compressed. The formula is F = -kx, where k is the spring constant and x is the displacement from the equilibrium position.

Can springs be used for more than just storing and releasing energy?

Yes, springs can also be used to control motion and absorb shock. They can be designed to dampen vibrations and reduce noise, making them useful in a variety of mechanical and industrial settings.

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