Two blocks of mass m1, m2 are attached to a spring

• Vriska
In summary: F}^{(1)} = -k(x_1+x_2)\hat{\jmath}## and ##\vec{F}^{(2)}=k(x_1+x_2)\hat{\jmath}##.##\vec{F}^{(1)} = -k(x_1+x_2)\hat{\jmath}## and ##\vec{F}^{(2)}=k(x_1+x_2)\hat{\jmath}##.That is the force on each body.What is the net force on each body?Now set them equal to ma. Then use the fact that in order for the system to do SHM, the net force
Vriska

Homework Statement

Now these two blocks are pulled separately by x1 and x2 distance. Find the frequency of shm

a = -w^2 x

The Attempt at a Solution

Now after m1 being pulled, it moves with acceleration
a1=Force of spring/m1.
acceleration of block 2 =
a2=/force of spring/m2.

Now accelerate the entire system in the direction of a1 because the frequency is independent of all extensions/accelerations.

so total acceleration of m2 = F/m1 + F/m2. This has effective mass 1/m1 + 1/m2.

So effective frequency would be square root of k/(1/m1 + 1/m2)

I'm kinda proud of this one tbh but I'm not sure if it's alright, so can someone tell me if this is sensible and I can use it in other places?

thanks

if I understand correctly you saying that: "In the non inertial frame of reference of m1, body m2 does SHM with effective mass ##\frac{m_1+m_2}{m_1m_2}##.

it can be tricky when working with non inertial frames but you might be correct here.

Are you actually able to show this result, step by step, starting from Newton's second law for each mass?

RedDelicious said:

Are you actually able to show this result, step by step, starting from Newton's second law for each mass?
-a1 m1 = F, a2 m2 = F. Turn on gravity (effective non inertial frame) on the direction of a2 with magnitude lf force equal a1m1. So force on block 1 is now -m1a1 +m1a1, force on m2 is a2m2 + a1m1= m2a... which is zero? wot.

so looks like you can't just add accelerations like velocities, ugh but why? this is a real big irregularity looks like.

btw how did you get the final answer? This questions leaving me stumped

Vriska said:
-a1 m1 = F, a2 m2 = F. Turn on gravity (effective non inertial frame) on the direction of a2 with magnitude lf force equal a1m1. So force on block 1 is now -m1a1 +m1a1, force on m2 is a2m2 + a1m1= m2a... which is zero? wot.

so looks like you can't just add accelerations like velocities, ugh but why? this is a real big irregularity looks like.

btw how did you get the final answer? This questions leaving me stumped

That will follow because the acceleration of the COM will be zero. There is no need to think about non-inertial reference frames. I can't think of anyway in which it doesn't just cause far more confusion than it's worth here.

Start fresh with a free body diagram.

What is the net force on m1?

What is the force on m2?

Now actually write this force out. Don't just leave it as F. Because they're connected by a spring, the force felt by either mass will depend on the position of either mass relative to the equilibrium positions. Do some tests to ensure that your forces makes sense at various positions for either mass. For example, F = 0 when x1=x2=0 and so forth.

Vriska said:
effective mass 1/m1 + 1/m2.

RedDelicious
haruspex said:
Strictly speaking you are correct but I guess he means with effective mass m such that ##\frac{1}{m}=\frac{1}{m_1}+\frac{1}{m_2}##.

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Delta² said:
Strictly speaking you are correct but I guess he means with effective mass m such that ##\frac{1}{m}=\frac{1}{m_1}+\frac{1}{m_2}##.
Perhaps, but the error found its way into the answer:
Vriska said:
So effective frequency would be square root of k/(1/m1 + 1/m2)
That would have dimension MT-1.

His initial reasoning isn't completely wrong.

Suppose F is the force from spring.

In an inertial frame, body m1 has acceleration ##-\frac{F}{m_1}## and body m2 ##\frac{F}{m_2}##.

in the non inertial frame of m1 , body m2 has acceleration ##\frac{F}{m_2}-\frac{-F}{m_1}=F(\frac{1}{m_1}+\frac{1}{m_2})## (and yes you can add accelerations as you can add velocities because the operator of derivative (with respect to time ) is linear).

His only mistake is that the effective mass is m such that ##\frac{1}{m}=\frac{1}{m_1}+\frac{1}{m_2}##.

And that the angular frequency of oscillation of the system is ##\sqrt{k(\frac{1}{m_1}+\frac{1}{m_2})}##.

Last edited:
Delta² said:
His initial reasoning isn't completely wrong.
I did not say it was. Just pointing out an algebraic slip that led to a wrong answer.

Delta2
RedDelicious said:
That will follow because the acceleration of the COM will be zero. There is no need to think about non-inertial reference frames. I can't think of anyway in which it doesn't just cause far more confusion than it's worth here.

Start fresh with a free body diagram.

What is the net force on m1?

What is the force on m2?

Now actually write this force out. Don't just leave it as F. Because they're connected by a spring, the force felt by either mass will depend on the position of either mass relative to the equilibrium positions. Do some tests to ensure that your forces makes sense at various positions for either mass. For example, F = 0 when x1=x2=0 and so forth.
alright I'm getting k(x1+x2( to be the force on both 1 and 2, how do i apply the acceleration of com constraints?

Delta² said:
His initial reasoning isn't completely wrong.

Suppose F is the force from spring.

In an inertial frame, body m1 has acceleration ##-\frac{F}{m_1}## and body m2 ##\frac{F}{m_2}##.

in the non inertial frame of m1 , body m2 has acceleration ##\frac{F}{m_2}-\frac{-F}{m_1}=F(\frac{1}{m_1}+\frac{1}{m_2})## (and yes you can add accelerations as you can add velocities because the operator of derivative (with respect to time ) is linear).

His only mistake is that the effective mass is m such that ##\frac{1}{m}=\frac{1}{m_1}+\frac{1}{m_2}##.

And that the angular frequency of oscillation of the system is ##\sqrt{k(\frac{1}{m_1}+\frac{1}{m_2})}##.

oh, yeah d/dt v1 + v2 = a1 + a2, i see.

but aren't non inertial frames just switching on gravity in one side? If these are equivalent, how do you get a1+a2 =a while doing this? I tried above but it bombed somewhere : /

and yah it should have been 1/m = 1/m1 + 1/m2, my bad

Vriska said:
oh, yeah d/dt v1 + v2 = a1 + a2, i see.

but aren't non inertial frames just switching on gravity in one side? If these are equivalent, how do you get a1+a2 =a while doing this? I tried above but it bombed somewhere : /
That's not the definition of non inertial frames (doesn't involve gravity as you say). From Wikipedia "A non-inertial reference frame is a frame of reference that is undergoing acceleration with respect to an inertial frame"

There must be some mistake in what you did because in the non intertial frame of m1, the total force on m2 can't be zero. If it was zero it would have acceleration zero in that frame. But the acceleration is not zero its ##F/m_1 + F/m_2## where F the force from the spring in the inertial frame. The mistake must be that in the non inertial frame of m1, the force from the spring to m2 is not the same as F.

Delta² said:
That's not the definition of non inertial frames (doesn't involve gravity as you say). From Wikipedia "A non-inertial reference frame is a frame of reference that is undergoing acceleration with respect to an inertial frame"There must be some mistake in what you did because in the non intertial frame of m1, the total force on m2 can't be zero. If it was zero it would have acceleration zero in that frame. But the acceleration is not zero its ##F/m_1 + F/m_2## where F the force from the spring in the inertial frame. The mistake must be that in the non inertial frame of m1, the force from the spring to m2 is not the same as F.

Isn't there something from Einstein that equates non inertial frames and gravity?

Vriska said:
Isn't there something from Einstein that equates non inertial frames and gravity?

I am not good with relativity theory sorry, so I can't comment on that.

One mistake I see when we consider the non inertial frame of m1, is that the spring c.o.m will have acceleration at this frame, so the two forces at the edges of the spring will not be equal and opposite as it happens in the inertial frame. They won't be F and -F as we consider them to be in the inertial frame.

But then again if we consider a massless spring...don't know how this case can be addressed.

Vriska said:
alright I'm getting k(x1+x2( to be the force on both 1 and 2, how do i apply the acceleration of com constraints?

That's not the right form. Draw a free body diagram. Also, the force won't be the same, only the magnitude will be. You need to keep track of your signs.

Vriska said:
Isn't there something from Einstein that equates non inertial frames and gravity?

No, definitely not. What you're thinking of is the equivalence principle but it does not equate non-inertial frames or inertial forces (also called pseudo forces) with gravity. "Switching on gravity" to indicate a non-inertial reference frame doesn't make any sense.

RedDelicious said:
That's not the right form. Draw a free body diagram. Also, the force won't be the same, only the magnitude will be. You need to keep track of your signs.
No, definitely not. What you're thinking of is the equivalence principle but it does not equate non-inertial frames or inertial forces (also called pseudo forces) with gravity. "Switching on gravity" to indicate a non-inertial reference frame doesn't make any sense.

Isn't force on m1 k(x1+x2) and force on m2 -k(x1+x2) because spring force my depends on total extension?

1. What is the equation for the spring force in this scenario?

The equation for spring force is F = -kx, where k is the spring constant and x is the displacement from the equilibrium position.

2. How do the masses affect the period of oscillation?

The period of oscillation is affected by the masses through the equation T = 2π√(m/k), where m is the total mass of the system and k is the spring constant.

3. How does the amplitude of oscillation change when the masses are changed?

The amplitude of oscillation is directly proportional to the masses. As the masses increase, the amplitude also increases, and vice versa.

4. What happens to the period of oscillation when the spring constant is increased?

As the spring constant increases, the period of oscillation decreases. This means that the system oscillates faster.

5. What is the equilibrium position in this system?

The equilibrium position is the point where the spring is at its natural length and there is no net force acting on the masses. This is the point where the system is at rest.

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