Spring SHM Homework: Solving for Displacement of a Suspended Point Mass

[Suyash Singh]

Homework Statement

A point mass m= 20 kg, is suspended by a
massless spring of constant 2000 N/m. The
point mass is released when elongation in
the spring is 15 cm. The equation of
displacement of particle as function of time
is : (Take g = 10 m/s2)

Homework Equations

A is amplitude
w is omega
t is time
k is spring constant

The Attempt at a Solution

PE= 1/2 m A^2 w^2 cos^2(wt+ phi)
PE=1/2Kx^2
equating these two equations doesn't do anything so how do i proceed?

 

[BvU]

doesnt do anything

Can you elaborate ? What is $\omega$ ? What is PE in the first equation ?

 

[haruspex]

equating these two equations doesn't do anything

It does if you consider specific values of x and t, i.e. the boundary conditions.
You will also need more equations since you have three unknowns to pin down.

 

[Suyash Singh]

s

Can you elaborate ? What is $\omega$ ? What is PE in the first equation ?

omega is angular velocity
PE is potential enrergy

 

[Suyash Singh]

It does if you consider specific values of x and t, i.e. the boundary conditions.
You will also need more equations since you have three unknowns to pin down.

i really don't know what to do

 

[Send BoBs]

So you want to find the equation of motion? Don't forget that the mechanical energy is being affected by gravity as well as the spring. So PE_gravity=mgx

You only used the kinetic energy and spring potential energy.

 

[haruspex]

i really don't know what to do

What do you know about x, $\dot x$ and P when t=0?

 

[Suyash Singh]

What do you know about x, $\dot x$ and P when t=0?

when t=0 x=15
dont know what is x dot(if it is omega then i don't know how to find it)
PE=1/2 2000 (15/100)^2
=1000 x 225/10000
22.5 joules

 

[Suyash Singh]

So you want to find the equation of motion? Don't forget that the mechanical energy is being affected by gravity as well as the spring. So PE_gravity=mgx

You only used the kinetic energy and spring potential energy.

but i don't know the distance from ground so PEgravity=??

 

[Send BoBs]

Here is what I would do. Just keep in mind that I only have the knowledge of a first year physics student

Particle executes SHM ∴
x(t)=x_mcos(ωt-θ)
v(t)=-ωx_msin(ωt-θ)
a(t)=-ω²x_mcos(ωt-θ)

F=ma => -1/2kx_m²=m(-ω²x_m) => ω=√k/m=10rad/s

x̶[̶S̶U̶B̶]̶m̶[̶/̶S̶U̶B̶]̶=̶0̶.̶0̶1̶5̶m̶, ω=10rad/s

v̶(̶0̶)̶/̶x̶(̶0̶)̶=̶-̶ω̶x̶[̶S̶U̶B̶]̶m̶[̶/̶S̶U̶B̶]̶s̶i̶n̶θ̶/̶x̶[̶S̶U̶B̶]̶m̶[̶/̶S̶U̶B̶]̶c̶o̶s̶θ̶=̶-̶ω̶t̶a̶n̶θ̶ ̶=̶>̶ ̶t̶a̶n̶θ̶=̶-̶(̶(̶v̶(̶0̶)̶/̶x̶(̶0̶)̶)̶/̶ω̶)̶ ̶=̶>̶ ̶θ̶=̶t̶a̶n̶[̶S̶U̶P̶]̶-̶1̶[̶/̶S̶U̶P̶]̶-̶(̶(̶v̶(̶0̶)̶/̶x̶(̶0̶)̶)̶/̶ω̶)̶ ̶s̶i̶n̶c̶e̶ ̶x̶(̶0̶)̶ ̶a̶n̶d̶ ̶v̶(̶0̶)̶ ̶a̶r̶e̶ ̶u̶n̶k̶n̶o̶w̶n̶ ̶t̶h̶i̶s̶ ̶m̶e̶t̶h̶o̶d̶ ̶f̶a̶i̶l̶s̶.̶ ̶ ̶∴̶x̶(̶t̶)̶=̶0̶.̶0̶1̶5̶m̶c̶o̶s̶(̶1̶0̶(̶r̶a̶d̶/̶s̶)̶t̶-̶θ̶)̶ ̶w̶h̶e̶r̶e̶ ̶o̶n̶l̶y̶ ̶θ̶ ̶i̶s̶ ̶l̶e̶f̶t̶ ̶u̶n̶k̶n̶o̶w̶n̶.̶ ̶T̶h̶i̶s̶ ̶i̶s̶ ̶a̶l̶l̶ ̶a̶s̶s̶u̶m̶i̶n̶g̶ ̶t̶h̶a̶t̶ ̶x̶[̶S̶U̶B̶]̶m̶[̶/̶S̶U̶B̶]̶=̶1̶5̶c̶m̶

There's a solution for this on https://www.hashlearn.com/questions/study_levels/class-xi/exams/engineering-entrance/subjects/physics/topics/oscillations-and-waves/challenges/point-mass-20-kg-suspended-massless-spring-constant-2000-nm-point-mass-released-elongat-aa4b9c6 but need to pay to see. If you're really desperate... I still wouldn't pay for that kind of information (should totally be free).

Also if you are using the mechanical energy the height for mgh would just be the distance x that is given when the particle is back at 0 potential spring energy and 0 when its first released.

 

[haruspex]

when t=0 x=15

First, need to define x. You need to specify where 0 is (bearing in mind that you have to match it with an SHM expression) and whether up is positive or negative.

dont know what is x dot

If x is displacement, what is dx/dt? What is its value when the mass is released?

 

[Suyash Singh]

Here is what I would do. Just keep in mind that I only have the knowledge of a first year physics student

Particle executes SHM ∴
x(t)=x_mcos(ωt-θ)
v(t)=-ωx_msin(ωt-θ)
a(t)=-ω²x_mcos(ωt-θ)

F=ma => -1/2kx_m²=m(-ω²x_m) => ω=√k/m=10rad/s

x_m=0.015m, ω=10rad/s

v̶(̶0̶)̶/̶x̶(̶0̶)̶=̶-̶ω̶x̶[̶S̶U̶B̶]̶m̶[̶/̶S̶U̶B̶]̶s̶i̶n̶θ̶/̶x̶[̶S̶U̶B̶]̶m̶[̶/̶S̶U̶B̶]̶c̶o̶s̶θ̶=̶-̶ω̶t̶a̶n̶θ̶ ̶=̶>̶ ̶t̶a̶n̶θ̶=̶-̶(̶(̶v̶(̶0̶)̶/̶x̶(̶0̶)̶)̶/̶ω̶)̶ ̶=̶>̶ ̶θ̶=̶t̶a̶n̶[̶S̶U̶P̶]̶-̶1̶[̶/̶S̶U̶P̶]̶-̶(̶(̶v̶(̶0̶)̶/̶x̶(̶0̶)̶)̶/̶ω̶)̶ ̶s̶i̶n̶c̶e̶ ̶x̶(̶0̶)̶ ̶a̶n̶d̶ ̶v̶(̶0̶)̶ ̶a̶r̶e̶ ̶u̶n̶k̶n̶o̶w̶n̶ ̶t̶h̶i̶s̶ ̶m̶e̶t̶h̶o̶d̶ ̶f̶a̶i̶l̶s̶.̶ ̶ ̶∴̶x̶(̶t̶)̶=̶0̶.̶0̶1̶5̶m̶c̶o̶s̶(̶1̶0̶(̶r̶a̶d̶/̶s̶)̶t̶-̶θ̶)̶ ̶w̶h̶e̶r̶e̶ ̶o̶n̶l̶y̶ ̶θ̶ ̶i̶s̶ ̶l̶e̶f̶t̶ ̶u̶n̶k̶n̶o̶w̶n̶.̶ ̶T̶h̶i̶s̶ ̶i̶s̶ ̶a̶l̶l̶ ̶a̶s̶s̶u̶m̶i̶n̶g̶ ̶t̶h̶a̶t̶ ̶x̶[̶S̶U̶B̶]̶m̶[̶/̶S̶U̶B̶]̶=̶1̶5̶c̶m̶

There's a solution for this on https://www.hashlearn.com/questions/study_levels/class-xi/exams/engineering-entrance/subjects/physics/topics/oscillations-and-waves/challenges/point-mass-20-kg-suspended-massless-spring-constant-2000-nm-point-mass-released-elongat-aa4b9c6 but need to pay to see. If you're really desperate... I still wouldn't pay for that kind of information (should totally be free).

Also if you are using the mechanical energy the height for mgh would just be the distance x that is given when the particle is back at 0 potential spring energy and 0 when its first released.

Its ok cause its a school problem
ok so omega, x are known now
i need phi and amplitude
x(t)=xsin(wt+ phi)
at t=0
15=15sin(phi)
phi=90 degrees
x(t)=15sin(10t+90)=15cos(10t)
similarly for others
v(t)=-150sin(10t)
a(t)=-1500cos(10t)

 

[Suyash Singh]

First, need to define x. You need to specify where 0 is (bearing in mind that you have to match it with an SHM expression) and whether up is positive or negative.

If x is displacement, what is dx/dt? What is its value when the mass is released?

x is displacement
x=0 is the condition when mass is not present on spring
down is positive cause its more convienient this way here
dx/dt is velocity
mass released then velocity=0 m/s cause of the above equation which i wrote in post #12

 

[haruspex]

x=0 is the condition when mass is not present on spring

For the standard SHM equation you need the displacement from the equilibrium position. Otherwise you will need to add a constant term.
So think again about the amplitude.

 

[Suyash Singh]

For the standard SHM equation you need the displacement from the equilibrium position. Otherwise you will need to add a constant term.
So think again about the amplitude.

but the three equations i got in post no 12 are they sufficient to find the solution?
is the amplitude 15cm?
but what is amplitude really here?
i mean not the value but the meaning of the term
in waves it is the up and down distance but here?

 

[haruspex]

what is amplitude really here?

In SHM the amplitude is the maximum displacement each side from a central equilibrium position.
Where is the equilibrium position in this case?

 

[Send BoBs]

If you draw a diagram of the motion of the mass you'll see the 15cm distance that the spring was stretched is greater than the distance from the equilibrium position to the fully stretched position. So the given distance x=15cm is greater than x_m.

But you do know the distance x=15cm so you may be able to solve for x_m with the mechanical energy at those two positions.

 

[Suyash Singh]

In SHM the amplitude is the maximum displacement each side from a central equilibrium position.
Where is the equilibrium position in this case?

If you draw a diagram of the motion of the mass you'll see the 15cm distance that the spring was stretched is greater than the distance from the equilibrium position to the fully stretched position. So the given distance x=15cm is greater than x_m.

But you do know the distance x=15cm so you may be able to solve for x_m with the mechanical energy at those two positions.

force=kx
mg/k=x
x=200/2000
x=0.1m

equilibrium position is 0.1 m

 

[haruspex]

Ok, so what is the initial displacement from equilibrium (i.e., the amplitude)?

 

[Suyash Singh]

Ok, so what is the initial displacement from equilibrium (i.e., the amplitude)?

0.15m-0.1m=0.05m
amplitude=0.05m

 

[haruspex]

0.15m-0.1m=0.05m
amplitude=0.05m

Right, so can you now write the complete expression? Don't forget to specify what your x represents.

 

[Suyash Singh]

Right, so can you now write the complete expression? Don't forget to specify what your x represents.

x(t)=0.05sin(10t)
x is displacement from equilibrium

 

[haruspex]

x(t)=0.05sin(10t)
x is displacement from equilibrium

Does that give the right displacement at t=0?

 

[Suyash Singh]

Does that give the right displacement at t=0?

displacement at t=0 is -0.1 m
oh :(
x(t)=0.05sin(10t) - 0.1

 

[haruspex]

displacement at t=0 is -0.1 m
oh :(
x(t)=0.05sin(10t) - 0.1

No, now you do not have it symmetric about x=0. Look back at your post #12. There is something there you have forgotten.
And it is the displacement from equilibrium position that is of interest. What is the initial displacement?

 

[Suyash Singh]

x(t)=Asin(wt+phi)
at t=0
displacement from equilibrium is 0.05m
so
0.05=0.05sin(phi)
thus phi=90 degrees
x(t)=0.05sin(10t+90)
x(t)=0.05cos(10t)

 

[haruspex]

x(t)=Asin(wt+phi)
at t=0
displacement from equilibrium is 0.05m
so
0.05=0.05sin(phi)
thus phi=90 degrees
x(t)=0.05sin(10t+90)
x(t)=0.05cos(10t)

Yes!