A Simple Harmonic Motion Question

In summary: Then t (period of the SHM) can be...In summary, the cylinder is displaced vertically by x and the buoyancy force increases by π*r2*x*1000*10.
  • #1

Homework Statement


Capture.JPG

As in the given picture, the cylinder is drowned (not completely drowned as in partially drowned) in water. The cylinder is attached with a spring which has the spring constant of 200 N/m. The spring has attached to a unmovable point in the ceiling. The weight of the cylinder is 5 kg and the weight of the spring is inconsiderate. If the cylinder pulled slightly down then will there be a simple harmonic motion and if so what is the Period?

Important note:
The height of the cylinder is not given.
The density of the cylinder is also not given.
The density of the water is 1000 kg/m^3.

Homework Equations



The hook's law -> F= -kx where F=force
k= the spring constant
x= is the displacement

The Attempt at a Solution


Capture.JPG
Capture.JPG


The cylinder is moving in accelerated motion. Let's take the acceleration as a.
Applying the Newton's Law:
F=ma regarding upward motion
Where,

F=T-50+U

ma= 5a

Where T= Tension
U= Up thrust

Hence:

T-50+U=5a

-200x-50+U=5a

U=Vdg

Where V= Volume of the cylinder under water
d= Density of the cylinder
g= gravitation acceleration

This is where I am stuck. I cannot identify the volume of the cylinder under water since it is changing during the motion. What should I do next...?
 
Physics news on Phys.org
  • #2
Are you not told the radius of the cylinder?
 
  • #3
haruspex said:
Are you not told the radius of the cylinder?
Didn't I mention it?
The radius is 6 cm
 
  • #4
Nipuna Weerasekara said:
Didn't I mention it?
The radius is 6 cm
Ok. When the cylinder is displaced vertically by x (upward positive?) what is the change in the buoyancy force?
 
  • #5
haruspex said:
Ok. When the cylinder is displaced vertically by x (upward positive?) what is the change in the buoyancy force?
That is the point where I can't understand any further. Please explain...
 
  • #6
Nipuna Weerasekara said:
That is the point where I can't understand any further. Please explain...
What is the change in the volume of water displaced? (If the radius of the beaker is not given then you will have to assume it is much wider than the cylinder.)
 
  • #7
How can I calculate it? when there is no height of the cylinder is given...?
 
  • #8
Nipuna Weerasekara said:
How can I calculate it? when there is no height of the cylinder is given...?
We only want the change in the submerged volume for a small change in x. The height of the cylinder will not matter.
 
  • #9
So that explains it...
But how can I identify the change in submerged volume?
Please elaborate...
 
  • #10
Nipuna Weerasekara said:
So that explains it...
But how can I identify the change in submerged volume?
Please elaborate...
If you push a vertical cylinder radius r into water to a depth of x, what volume is submerged?
 
  • #12
Nipuna Weerasekara said:
That would be
Volume = π*r2*x
Then what?
Right. If the submerged volume increases by that much, how much does the buoyancy increase by?
 
  • #13
U = Vdg
U = π*r2*x*1000*10
U = 10000πr2x

Then what?
 
  • #14
Nipuna Weerasekara said:
U = Vdg
U = π*r2*x*1000*10
U = 10000πr2x

Then what?
It's better not to plug in numbers until the end. For now, keep it as ##A\rho_wgx##, where A is the cross-sectional area of the cylinder, i.e. ##\pi r^2##.
Next is to write out the force and acceleration equation for the vertical movement of the cylinder. If the cylinder is height x above its equilibrium position, what is the net force acting on it?
 
  • Like
Likes Nipuna Weerasekara
  • #15
What's the radius/diameter of the beaker?
 
  • #16
Khashishi said:
What's the radius/diameter of the beaker?
See post #3
 
  • #17
The radius of the beaker affects how much the water level moves up and down when the cylinder moves.

Edit: I guess it doesn't matter in the end.
 
Last edited:
  • Like
Likes Nipuna Weerasekara
  • #18
Khashishi said:
The radius of the beaker affects how much the water level moves up and down when the cylinder moves.

Edit: I guess it doesn't matter in the end.
Sorry, I should have said "see post #6". Yes, I agree that the radius of the beaker matters too, but if it's not given we will have to take it as being effectively infinite.
 
  • #19
The problem was solved... The radius of the beaker does not matter nor the height of the cylinder and the density of it.
The method to find the answer is first we have to draw a step of equilibrium of the cylinder.
Then we have to note that the submerged volume of cylinder is h at the equilibrium stage. Whereas the displacement of the spring is noted as c.
Then we can see
F = -kx
F = T
x = c
T = -kc
U = Ahdg
Considering equilibrium of the cylinder:
F= ma
T + U -mg = 0
-kc + Ahdg -mg =0

Then we go to the step where we pull the cylinder for x distance
Then,
Tx = -k (c+x)
F= ma for up
Tx + Ux - mg = ma
Ux = A(h+x)dg
-k (c+x) + A(h+x)dg - mg = ma

-kc -kx + Ahdg +Axdg -mg =ma

-kx + Axdg = ma

The final equation is

a = -(Adg - k)/m * x

Hence the ω2 = (Adg - k)/m
Then ω2 = 2π/t
A = π*0.062 m2 = 0.0036π m2
d = 1000 kgm-3
g = 10 ms-2
k = 200 Nm-1
m = 5 kg
Then t (period of the SHM) can be found...

Thanks...
 
Last edited:
  • #20
Nipuna Weerasekara said:
The radius of the beaker does not matter
It does. Suppose the cylinder has radius r and the beaker has radius R. If the cylinder is displaced upward by x then the spring tension reduces by kx, but the volume of cylinder submerged reduces by ##\pi x\frac{R^2r^2}{R^2-r^2}##.
 
  • Like
Likes Nipuna Weerasekara
  • #21
Yes that is true but if we use it in this calculation we cannot come to an exact solution like I have came. By the way my teacher said that I should not consider the change in water levels because in this level I don't have to use it in the calculations.
By the way thanks for the valuable advises...
 
  • #22
Nipuna Weerasekara said:
Yes that is true but if we use it in this calculation we cannot come to an exact solution like I have came. By the way my teacher said that I should not consider the change in water levels because in this level I don't have to use it in the calculations.
By the way thanks for the valuable advises...
Ok.
 
  • Like
Likes Nipuna Weerasekara

What is simple harmonic motion?

Simple harmonic motion is a type of periodic motion in which an object oscillates back and forth around a central point, and its displacement from that point is directly proportional to the force acting on it.

What are the characteristics of simple harmonic motion?

The characteristics of simple harmonic motion include a constant amplitude, equal time intervals between oscillations, and a sinusoidal (or circular) motion. It also has a period (time for one complete cycle) and frequency (number of cycles per second).

What is the equation for simple harmonic motion?

The equation for simple harmonic motion is x(t) = A * cos(ωt + φ), where x is the displacement, A is the amplitude, ω is the angular frequency, and φ is the phase angle.

What factors affect simple harmonic motion?

The factors that affect simple harmonic motion include the mass of the object, the force acting on it, and the stiffness of the spring or medium that the object is moving in.

What are some real-life examples of simple harmonic motion?

Some real-life examples of simple harmonic motion include the motion of a pendulum, the vibration of a guitar string, and the motion of a mass attached to a spring.

Suggested for: A Simple Harmonic Motion Question

Replies
5
Views
898
Replies
6
Views
2K
Replies
5
Views
945
Replies
4
Views
773
Replies
1
Views
1K
Replies
8
Views
1K
Back
Top