Motion of box on inclined plane connected by spring to a wall

[haruspex]

I get:
$$T=mg \sin \alpha + ma \cos \alpha - ma$$

and
$$a=\frac{mg \sin \alpha -kx}{M + 2m(1-\cos \alpha)}$$

How to proceed to find period? Thanks

Remember that a is $\ddot x$. The standard form of the SHM ODE is $\ddot x+\omega^2x=0$, but that is taking x as offset from the equilibrium position. Because your x is not measured from there the form you have is $\ddot x+\omega^2x=C$.
You could use the equilibrium position you found to adjust your x to be an offset from there. If you have done everything correctly c should disappear. Or you could just trust that it will disappear and ignore it.

 

[songoku]

Remember that a is $\ddot x$. The standard form of the SHM ODE is $\ddot x+\omega^2x=0$, but that is taking x as offset from the equilibrium position. Because your x is not measured from there the form you have is $\ddot x+\omega^2x=C$.
You could use the equilibrium position you found to adjust your x to be an offset from there. If you have done everything correctly c should disappear. Or you could just trust that it will disappear and ignore it.

Oh, so $\omega^2$ will be $\frac{k}{M+2m(1-\cos \alpha)}$ then using $\omega=\frac{2 \pi}{T}$ we can find the period

Sorry I have several questions:

I don't think that follows. Remember that a_mw is down and left.

Why can't I say $m$ will move to the right with respect to the ground? My reasoning is since the horizontal acceleration of $m$ is $a-a \cos \alpha$ and $|a|>|a \cos \alpha|$ and direction of $a$ is to the right, so horizontal acceleration of $m$ is to the right.

Or do you mean the direction of motion of $m$ with respect to the ground should be the resultant of horizontal and vertical direction?

For question (d), would it be like, for M:
$$\Sigma F=M.a$$
$$N \sin \alpha - k.x = M.\frac{v.dv}{dx}$$
$$\int_0^{x_0} (N \sin \alpha - k.x)dx = M \int_p^{q} v~dv$$

Is this correct? And I suppose the value of $q$ = 0 but what will be the value for $p$?

Thanks

Is this valid approach for (d)? If yes, what would be the limit for the RHS integration?

For (e), I tried to use conservation of energy again but got stuck

Let:
$d$ = distance move by $M$ from initial position to reach equilibrium point
$v_m$ = speed of $m$ after moving down the slope as far as $d$
$v_M$ = speed of $M$ after moving a horizontal distance $d$ to the right
Ground level is the position of $m$ after moving down a distance $d$ down the slope

I compare the initial condition where the system is at rest and when $M$ reaches equilibrium point

$$mg~d \sin \alpha = \frac1 2 m (v_m)^2 + \frac 1 2 M (v_M)^2 + \frac 1 2 k d^2$$

$v_m$ and $v_M$ will be the maximum speed of the oscillation so $v_m = v_M = \omega A$ where $A$ is the amplitude of oscillation and $A=x_0-d$

So:
$$mg~d \sin \alpha = \frac1 2 m (v_m)^2 + \frac 1 2 M (v_M)^2 + \frac 1 2 k d^2$$
$$mg~d \sin \alpha=\frac 1 2 m (\omega (x_0-d))^2 + \frac 1 2 M (\omega (x_0-d))^2 + \frac 1 2 kd^2$$

How to get rid $\omega$ from the equation? Thanks

Can I change $\omega$ in the last equation with $\omega = \sqrt{\frac{k}{M+m}}$

I think these types of problems where there is a motion within another motion or multiple motions relative to ground unfolding in a complex manner, it's best to use the accelerating object as a frame of reference i.e. a non-inertial frame of reference to simplify things and then add a pseudo force on the other object being observed from an accelerating frame of reference.

In this situation, let's have an observer on accelerating wedge. This observer will see the block m as stationary since it's tied to the wall with a fixed length string. If a is the initial acceleration of wedge then apply a force ma in opposite direction on block m. We are focusing on the instant t=0 since we need the acceleration of wedge at t=0.

Below are the FBDs that I get for block m relative to accelerating wedge and for wedge relative to ground. Note that at t=0 there is no spring force since the problem states that the spring is not stretched initially. The second FBD is relative to ground, while first is relative to wedge.

Block observed from wedge
View attachment 281133
View attachment 281134

Where can I learn about non-inertial frame in more detail?

Thanks

 

[haruspex]

Why can't I say m will move to the right with respect to the ground?

You can. I was handling two threads at once which had remarkably similar problems. m would have moved left in the other one.

Is this valid approach for (d)?

Yes, but you forgot the force from the string on the pulley again.
p and q would be the initial and final speeds, which are both zero here.
You should get the same equation you got in post #46.

 

[haruspex]

Can I change $\omega$ in the last equation with $\omega = \sqrt{\frac{k}{M+m}}$

No. Using KE was harder than I thought. Stick with your force approach.

 

[songoku]

Thank you very much for all the help and explanation haruspex, Lnewqban, Tsny, etotheipi, vcsharp2003

 

[vcsharp2003]

Where can I learn about non-inertial frame in more detail?

Its really very simple as explained below. You can look up https://en.wikipedia.org/wiki/Fictitious_force for more details.

If you put an observer in an accelerating frame of reference then the object being observed will have all "real" forces acting on it like gravity, tension, friction, normal reaction etc. PLUS a pseudo force that equals ma in magnitude having a direction opposite to the acceleration of frame of reference where a is acceleration magnitude of reference frame and m is mass of object being observed . We can then apply the Newtons Second Law of Motion just as we normally do in a non-inertial frame of reference. This is a standard practice when solving complex problems where one of the object is moving in a complex and hard-to-predict manner; the pseudo force simplifies our analysis of the problem greatly. In my view, the problem that you posted under this thread is an ideal candidate for such an analysis since it makes it very simple as the smaller mass is in equilibrium with respect to the wedge.

A common area of application for pseudo force is in uniform circular motion where the object in circular motion has a radial acceleration towards the center of the circle. If we now put an observer in a frame of reference that is having the same circular motion as the object then we have our observer in an accelerating frame of reference and the object will appear at rest relative to the frame of reference. We apply a force ma = mv²/r in a radially outward direction. This radial outward force is called centrifugal force and it's an example of a pseudo force.

Pseudo force is a fictitious force which has no real existence, but it helps us to apply Newton's Second Law of Motion from an accelerating frame of reference. As a rule, Newton's Second Law of Motion can only be applied when observer is in an inertial frame of reference (i.e. frame is at rest or moving at a constant velocity relative to ground).

 

[songoku]

Thank you again vcsharp2003 for al the explanation