Show that the simple harmonic motion has a period

[Myr73]

A mass m slides on a frictionless horizontal surface, connected to two springs, as shown below. If the force have constants K1 and K2 , show that the simple harmonic sliding motion has period.
T= 2 Pie (Square Root X m { K1+K2/K1K2})

So far I have →From equation T= 2Pie{SquareRoot(m/k)
If these two equations are the same, k must equal { K1+K2/K1K2}.

It may just be the calculus that I am not understanding but I don't know how to show that
k= { K1+K2/K1K2}. How would I begin this?

 

[rude man]

You should start by writing F = ma equation for the mass as a function of displacement from a zero point.

 

[Myr73]

Sorry, I am still unsure what to do here.

 

[rude man]

Displace the mass a distance away from the x=0 position (where both springs are relaxed). What is the force on the mass?

PS - where's the picture?

 

[Myr73]

Sorry the picture doesn't work. It is K1 spring on the left, attached to k2 spring , which is attached to m at the far right. All horizontal

 

[rude man]

Sorry the picture doesn't work. It is K1 spring on the left, attached to k2 spring , which is attached to m at the far right. All horizontal

OK, so back to the question in post 4 ...

Wait a minute - the mass isn't between the two springs?
So, let x=0 be where the two springs are relaxed, then displace the mass an amount x= +x0, what is the force on the mass?

 

[collinsmark]

A mass m slides on a frictionless horizontal surface, connected to two springs, as shown below. If the force have constants K1 and K2 , show that the simple harmonic sliding motion has period.
T= 2 Pie (Square Root X m { K1+K2/K1K2})

So far I have →From equation T= 2Pie{SquareRoot(m/k)
If these two equations are the same, k must equal { K1+K2/K1K2}.

k₁ does not need to be able to k₂ for this to work. But you will soon find out that there are certain things that do turn out to be equal.

It may just be the calculus that I am not understanding but I don't know how to show that
k= { K1+K2/K1K2}. How would I begin this?

Start with rude man's advice. That's a good approach.

Keep things static for now (don't worry about motion or acceleration just yet -- we'll get back to that). So the first step is to answer rude man's question in post #6.

The answer to that might not be so obvious, which is why I'm chiming in here. What does Newton's third law of motion say about the relationship is between the force on the mass and the force on spring k₂?

Invoke Newton's third law again, regarding the force on spring k₁.

Then realize one final constraint. What is the total displacement of both springs k₁ and k₂ added together?

 

[rude man]

Or, how about this: combine the 2 springs into one spring:

first, find the net spring constant k: 3 equations, 3 unknowns: stretch of spring 1, stretch of spring 2, effective k of the series-connected springs.

Then, you have 1 spring & 1 mass to contend with.

 

[Myr73]

Ok, umm Force on mass at x=+x0 --> Third law;For every action there is an equal and opposite reaction--> Therefore Force on mass is opposite to Force on Springs. Is that what you meant?... I believe I may need more explanation on what to do with this.

 

[ehild]

The two springs are connected in series, and the combined spring acts on the mass.

What is the spring constant of two springs connected in series? See picture.

You pull the orange spring with force F. At equilibrium, the net force for both springs is zero: the orange spring is pulled by F and at the other end it is pulled backwards by the blue spring also by F. The orange springs pulls the blue one with F, and the wall pulls the blue spring backwards with the same force. The tension is the same along a spring.

Both springs stretch, the blue one with Δx1, the red one with Δx2. ##Δx_1=\frac{F}{k_1}## ,##Δx_2=\frac{F}{k_2}##.
The whole spring stretches by ΔX=Δx₁+ Δx₂ by the force F. The resultant spring constant of the connected springs is [tex]K=\frac{F}{ΔX}=\frac{F}{Δx_1+Δx_2}=\frac{F}{\frac{F}{k_1}+\frac{F}{k_2}}=\frac{k_1k_2}{k_1+k_2}[/tex]

ehild

 

[rude man]

Ok, umm Force on mass at x=+x0 --> Third law;For every action there is an equal and opposite reaction--> Therefore Force on mass is opposite to Force on Springs. Is that what you meant?... I believe I may need more explanation on what to do with this.

If you can handle the one-spring + mass problem you can handle this one the same way - see my post 8 again.

 

[collinsmark]

Ok, umm Force on mass at x=+x0 --> Third law;For every action there is an equal and opposite reaction--> Therefore Force on mass is opposite to Force on Springs. Is that what you meant?...

Well, yes, but you can break it down further (ehild's picture and post demonstrate this).

The mass is not in contact with the "springs," as in plural, but rather the mass is only in contact with spring k₂. What does Newton's third law of motion say about the force on spring k₂ and the force acting on the mass? (Note that nothing else is acting on the mass besides spring k₂).

Similarly, what is the compressive force of spring k₁ relative to that of spring k₂? You can use Newton's third law to determine this at the connection point between the two springs.

That's what I meant about finding several things which are equal.

The individual, net displacement of each spring is not equal though. The springs will most certainly be compressed by different displacements, if the forces on the springs are equal and spring constants k₁ and k₂ are not equal.

But there is a constraint. If you add the individual, net displacement of spring k₁ to the displacement of spring k₂, how does that relate to the displacement of the mass?

[Edit: the above approach is valid if the springs are massless, themselves.]

 

[Myr73]

Thank you.. It all makes great sense,however there is one thing I am not sure of and its the switch from
F/ F/k1+F/k2 to k1 k2/k1+k2

 

[collinsmark]

Thank you.. It all makes great sense,however there is one thing I am not sure of and its the switch from
F/ F/k1+F/k2 to k1 k2/k1+k2

For reference,

[tex] \frac{F}{\frac{F}{k_1} + \frac{F}{k_2}} [/tex]

As a first, interim step, find a common denominator of the [itex] \frac{F}{k_1} + \frac{F}{k_2} [/itex] terms in the overall denominator.

 

[ehild]

Thank you.. It all makes great sense,however there is one thing I am not sure of and its the switch from
F/ F/k1+F/k2 to k1 k2/k1+k2

Factor out F from the denominator, and simplify.
[tex]\frac{F}{\frac{F}{k_1} + \frac{F}{k_2}}=\frac{F}{F(\frac{1}{k_1}+\frac{1}{k_2})}=\frac{1}{\frac{1}{k_1}+\frac{1}{k_2}}[/tex]
Then do what colinsmark suggested, find the common denominator of 1/k1+1/k2...

ehild