# Physics problem relating to an inclined plane and a spring oscillation

• arhzz
In summary: In using ##F = kl##, ##F## is the force that stretches the cable a distance ##l## and holds it there. If you are going to use ##l = 15 ##cm, then you must use the the corresponding force. You are correct.
arhzz
Homework Statement
In a mine shaft inclined at 30 ° to the horizontal, three 10-tonne trucks, coupled to one another, are held by a steel cable. The steel cable is elastically stretched by 15 cm compared to the unloaded state. Suddenly the last of the three trolleys comes loose. Calculate the oscillation duration (period) and amplitude of the oscillation of the remaining two carts.
Relevant Equations
$$T = \pi \cdot 2 \sqrt \frac l g$$
Hello!

So my main and first problem about this question is, I do not know what the problem is about. What I mean by that is, in class we talked about pendulums and are given formulas and assignments regarding pendulums. But this problem here does not seem like it has anything to do with pendulums. Maybe I am just missing the obvious but it is nothing compared to like the other problems we are given.So I've tried going the "blindfolded way" simply follow the formulas to the solutions. In my lecture notes I've found this formula for the oscillation period

$$T = 2 \cdot \pi \sqrt \frac l g$$

T = 0,77 s

Now for the amplitude I have not found a formula for,and I honestly am not comftorable doing this problem until I know what I'm dealing with.

Thank you!

You're dealing with a mass-spring system.

gneill said:
You're dealing with a mass-spring system.
Okay, never heared of that,but thank you now I finally know what to do. Thank you!

Okay so I've done some reaserch and this is what I've come up with. So the formula to get the Period is this

$$T = 2 \cdot \pi \sqrt \frac m k$$

Now k we get like this,

$$k=\frac F l$$
$$k=\frac {m\cdot g} l$$
Since we are looking only for the period of the 2 remaining wagons our mass is 20t, so after calculating k with 20t we get this

k = 130800

If we plug it in the formula for T we get;

T = 0,776 s

Now I need to get the value of the amplitude,but could anyone confirm that my calculations make even a bit of sense now?

arhzz said:
$$k=\frac {m\cdot g} l$$
Did you take into account the slope of the mine shaft? Should ##m## be the mass of two trolleys or three trolleys? Draw a free-body diagram for the equilibrium situation when all three trolleys are on the end of the cable.

TSny said:
Did you take into account the slope of the mine shaft? Should ##m## be the mass of two trolleys or three trolleys? Draw a free-body diagram for the equilibrium situation when all three trolleys are on the end of the cable.
Well I did not take the slope into account, to be honest I'm not sure what to do with it because I haven't found anywhere the formula regarding period that the angle is taken into account. Now the mass should be two because the question states " Calculate the oscillation duration (period) and amplitude of the oscillation of the remaining two carts. " So I'd reckon it is only two, what makes you think otherwise? Now for the diagramm, I made my self a litle picture on how the problem looks like, but what am I susposed to see from it?

You are unlikely to easily find a ready-made formula for this particular situation. Most problems require you to analyze the situation and apply basic principles to build up your own formulas.

You will need to develop a strategy that combines what you know about forces acting on masses on ramps and Hooke's law for springs.

arhzz said:
So I'd reckon it is only two, what makes you think otherwise?
In using ##F = kl##, ##F## is the force that stretches the cable a distance ##l## and holds it there. If you are going to use ##l = 15 ##cm, then you must use the the corresponding force. Note that the force stretching the cable is not the entire weight of the trolleys but only a certain component of the weight. This is why you need a good force diagram.

TSny said:
In using ##F = kl##, ##F## is the force that stretches the cable a distance ##l## and holds it there. If you are going to use ##l = 15 ##cm, then you must use the the corresponding force. Note that the force stretching the cable is not the entire weight of the trolleys but only a certain component of the weight. This is why you need a good force diagram.
Okay now I am confused, the force F is stretching the cable to 15 cm that is obvious but " Note that the force stretching the cable is not the entire weight of the trolleys but only a certain component of the weight. " the way I interpret this is that not all 3 trolleys are stressing the cable, meaning not all of them are part of the force that is streching the cable. If that is so than my assumption that only 2 of the trollys are going into the calculation of F. Or are you implying that 2 trolleys are holding it at 15 cm but the third trolley weight is what makes the cable actually snap? Than again how does the angle play into all of this. I'd assume he has something to do with the forces,but now this with the weight is kind of messing with my head.

Initially, you have three trolleys that are held at rest by the cable and cause the cable to be stretched 15 cm.

TSny said:
Initially, you have three trolleys that are held at rest by the cable and cause the cable to be stretched 15 cm.
Okay I've read through your post carefully and now I see what you are implying. I've done my calculations and I'm too tired to post the entire process but here are the final results

T = 0,63 s
Amplitude = 0,5 m

EDIT: Okay I've been looking at my calculation a bit an I've stumbled upon a dilema;

I've used this formula to get T;

##\omega = \sqrt \frac {k} {m \cdot sin(30)} ##

And ##\omega = \frac {2\pi} T##

Not T from this is

##T = 2\pi \sqrt \frac {m \cdot sin(30)} k##

T = 0,63s

To get k I've used 30t and to get T I've used 20t.

But what crossed my mind is that I don't actually need to use the sin in the equation for T. An equation for T already exists ,and I've used it before.

##T = 2\pi \sqrt \frac m k##

And after I try it with this formula I get T = 0,897s and I'm not sure which is right? What would I have to spot to see which equation to use.

Last edited:
arhzz said:
##T = 2\pi \sqrt \frac m k##

And after I try it with this formula I get T = 0,897s and I'm not sure which is right? What would I have to spot to see which equation to use.
I believe this is correct. In the formula ##T = 2\pi \sqrt \frac m k##, ##m## represents the amount of inertia on the end of the "spring". So, ##m## is just the mass. There would not be a factor of ##\sin \theta## here.

I also think your answer of 5 cm for the amplitude is correct.

arhzz
TSny said:
I believe this is correct. In the formula ##T = 2\pi \sqrt \frac m k##, ##m## represents the amount of inertia on the end of the "spring". So, ##m## is just the mass. There would not be a factor of ##\sin \theta## here.

I also think your answer of 5 cm for the amplitude is correct.
Thank you! I've thought about it as well and to be honest I was leaning more towards the solution without the sin.

Good. I just noticed that earlier you wrote the amplitude as 0.5 m. Did you mean 0.05 m?

arhzz
TSny said:
Good. I just noticed that earlier you wrote the amplitude as 0.5 m. Did you mean 0.05 m?
Oh yes its 0,05 I've divided by 100 but forgot to add the 0.

TSny

## 1. How does the angle of inclination affect the oscillation of a spring on an inclined plane?

The angle of inclination plays a crucial role in the oscillation of a spring on an inclined plane. The steeper the angle, the greater the force of gravity acting on the spring and the faster the oscillation will be. Conversely, a smaller angle will result in a slower oscillation due to a weaker force of gravity.

## 2. What is the relationship between the spring constant and the amplitude of oscillation on an inclined plane?

The spring constant, also known as the stiffness of the spring, determines the frequency of oscillation on an inclined plane. A higher spring constant will result in a higher frequency and a smaller amplitude, while a lower spring constant will lead to a lower frequency and a larger amplitude.

## 3. How does the mass of the object on the inclined plane affect the oscillation of the spring?

The mass of the object on the inclined plane has a direct impact on the oscillation of the spring. A heavier object will result in a stronger force of gravity, leading to a faster oscillation. On the other hand, a lighter object will result in a weaker force of gravity and a slower oscillation.

## 4. What is the role of friction in an oscillating spring on an inclined plane?

Friction plays a significant role in an oscillating spring on an inclined plane. It acts in the opposite direction of the motion, slowing down the oscillation. The steeper the angle of the inclined plane, the greater the friction, and the slower the oscillation will be.

## 5. How does the length of the spring affect the period of oscillation on an inclined plane?

The length of the spring is directly proportional to the period of oscillation on an inclined plane. A longer spring will have a longer period, while a shorter spring will have a shorter period. This is because a longer spring has a lower spring constant, resulting in a slower oscillation.

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