# Prove that a mass has simple harmonic motion

#### Like Tony Stark

Homework Statement
A body of mass $10 kg$ is connected to a spring of constant $490 \frac{N}{m}$ and it lies on an frictionless inclined plane. The angle formed by the plane and the floor is $30°$. This block is inside a room which accelerates upwards with $a=5 \frac{m}{s^2}$. Prove that if the mass is moved away from its equilibrium point, it will experience simple harmonic motion
Homework Equations
Newton's equation
If I write Newton's equations, seen inside the room and with non tilted axis we have:
$x) N.sin(\alpha)-Fe.cos(\alpha)=m.a_x$
$y) N.cos(\alpha)+Fe.sin(\alpha)-m.g-f*=m.a_y$

Where $f*=ma$, $Fe$ is the elastic force.

Then, how can I realize about simple harmonic motion?

I also can think it with tilted axis, which would be
$x) mg.sin(\alpha)+f*sin(\alpha)-Fe=m.a_x$
$y)N-mg.cos(\alpha)-f*cos(\alpha)=0$

But I can't notice the SHM. I mean, I can't relate that with $m.\ddot x +k.x=0$

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#### Orodruin

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Your quoted differential equation for SHM is valid only when x is the displacement from equilibrium. You need to find the equilibrium point and expand your differential equations around it.

#### PeroK

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Homework Statement: A body of mass $10 kg$ is connected to a spring of constant $490 \frac{N}{m}$ and it lies on an frictionless inclined plane. The angle formed by the plane and the floor is $30°$. This block is inside a room which accelerates upwards with $a=5 \frac{m}{s^2}$. Prove that if the mass is moved away from its equilibrium point, it will experience simple harmonic motion
You have a) gravity; b) the room accelerating; c) an incline; d) a mass on a spring.

What is the net result of the combination of a) and b)?

Is this fundamentally different from the normal situation where the room is not accelerating?

Does a mass on a spring on a flat surface exhibit SHM? If so, why?

Does a mass on a spring hanging vertically exhibit SHM? If so, why?

In what way is a mass on a spring on a inclined plane different from the vertical case? In what way is it the same?

#### Like Tony Stark

You have a) gravity; b) the room accelerating; c) an incline; d) a mass on a spring.

What is the net result of the combination of a) and b)?

Is this fundamentally different from the normal situation where the room is not accelerating?

Does a mass on a spring on a flat surface exhibit SHM? If so, why?

Does a mass on a spring hanging vertically exhibit SHM? If so, why?

In what way is a mass on a spring on a inclined plane different from the vertical case? In what way is it the same?
If you compare this case with the simplest case, now you have a component of the weight and a component of the pseudo-force. So you have two more forces. And that's the difference, if it wasn't inclined there wouldn't be weight (at least, It wouldn't matter) and if it wasn't accelerated there wouldn't be a pseudo force.
(The forces on $y$ will be 0)

#### Like Tony Stark

Your quoted differential equation for SHM is valid only when x is the displacement from equilibrium. You need to find the equilibrium point and expand your differential equations around it.
How? I mean, suppose that the equilibrium point is $l_eq$, then what should I do? How do I have to add the weight and pseudo force in the differential equation?

#### PeroK

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If you compare this case with the simplest case, now you have a component of the weight and a component of the pseudo-force. So you have two more forces. And that's the difference, if it wasn't inclined there wouldn't be weight (at least, It wouldn't matter) and if it wasn't accelerated there wouldn't be a pseudo force.
(The forces on $y$ will be 0)
If someone were to suggest that in all cases the SHM would be defined by the spring constant, the mass and the initial displacement from equilibrium, then this would be:

1) So obviously nonsensical that it's not worth thinking about? The scenarios are clearly fundamentally different?

2) An interesting idea that might stimulate a deeper analysis? Maybe the scenarios are not so different as they at first appear?

#### Like Tony Stark

If someone were to suggest that in all cases the SHM would be defined by the spring constant, the mass and the initial displacement from equilibrium, then this would be:

1) So obviously nonsensical that it's not worth thinking about? The scenarios are clearly fundamentally different?

2) An interesting idea that might stimulate a deeper analysis? Maybe the scenarios are not so different as they at first appear?
It's not that I'm not trying to study it, but I don't understand what I should see for being able to say "Oh! This is simple harmonic motion". That's why I quoted the differential equation, I thought that that's the only way to determine if a motion it's SHM

#### PeroK

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It's not that I'm not trying to study it, but I don't understand what I should see for being able to say "Oh! This is simple harmonic motion". That's why I quoted the differential equation, I thought that that's the only way to determine if a motion it's SHM
Okay, my questions were rhetorical. To be clear: I'm telling you that the problem can be greatly simplified by identifying irrelevant factors. You've written down some equations that include everything, but you should simplify the problem first.

Also, for example, you have written down components of motion along x and y axes. But, motion of the mass is constrained along the incline. So, taking the incline as the axis along which motion occurs is a good simplifying step. It reduces a 2D problem to a 1D problem.

You might also think about what equilibrium means. It means, in this case, that all forces relative to the accelerated reference frame are balanced.

Finally, you may or may not need to show that the simple case of a mass on a spring exhibits SHM.

#### Orodruin

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How? I mean, suppose that the equilibrium point is $l_eq$, then what should I do? How do I have to add the weight and pseudo force in the differential equation?
You do not suppose that equilibrium is somewhere. You compute where it is - it is where the net force on the block is zero.

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