## Given a cart and spring on incline, how do I find the period of oscillation?

Turnnirl
Posts: 1
Joined: Mon Oct 15, 2018 7:00 pm

### Given a cart and spring on incline, how do I find the period of oscillation?

hi I need a question answered

I need help on b

A cart of mass m is attached to a vertical spring of force constant k so that the spring stretches 20 cm. When the cart is set into oscillatory motion on the vertical spring, the period of the oscillation is 3 seconds. The cart is set on an incline angled at 30 and is attached to the top of the plane by the by the same spring.
(a) What is the stretch of the spring while the cart is on the incline?
(b) What is the period of the oscillatory motion that the cart could undergo on the incline.

For part a I got 10 cm and it was marked correct.

Here's where I'm not sure I'm going in the correct direction. To put in gravity, I introduce an angle theta. But I need to know more to use it.

jeff
Posts: 38
Joined: Mon Mar 05, 2018 11:54 pm

### Re: Given a cart and spring on incline, how do I find the period of oscillation?

Hi Turnnirl,
Turnnirl wrote:
Mon Oct 15, 2018 7:00 pm
hi I need a question answered

I need help on b

A cart of mass m is attached to a vertical spring of force constant k so that the spring stretches 20 cm. When the cart is set into oscillatory motion on the vertical spring, the period of the oscillation is 3 seconds. The cart is set on an incline angled at 30 and is attached to the top of the plane by the by the same spring.
(a) What is the stretch of the spring while the cart is on the incline?
(b) What is the period of the oscillatory motion that the cart could undergo on the incline.

For part a I got 10 cm and it was marked correct.

Here's where I'm not sure I'm going in the correct direction. To put in gravity, I introduce an angle theta. But I need to know more to use it.
It turns out that because gravity is a constant force (near earth) and the spring force is a linear force (in displacement) that the force of gravity actually has no effect on the period in these cases. For example, when the spring is horizontal, like on a floor, the mass oscillates about the unstretched spring position. When it is vertical, gravity causes the mass to oscillate about a different point, where the spring is already stretched. But the period of motion about that point is the same.

And on the incline, it's inbetween. Gravity causes the spring's "neutral" position (the "rest" position) to be farther down, but not as far down as the vertical. But still, the period is the same in all 3 cases, because that's just the motion that the spring causes back and forth about the "rest" position.

Does that make sense? The spring will stretch enough to cancel the force of gravity, and then can oscillate about that new point. It can be a tricky point to see how they cancel while the spring length is moving, so please ask if it's not clear.

roxieqk16
Posts: 2
Joined: Fri Nov 16, 2018 11:50 pm

### Re: Given a cart and spring on incline, how do I find the period of oscillation?

I don't understand your answer to the OP. For springs I can find the differential equation for a horizontal and vertical spring, but how can they be the same? The equations are

m d^2x/dt^2 + k x = 0

m d^2x/dt^2 + k x + m g = 0

How can those give the same results whenthe equation is different? And also the same when the spring is on an incline and the equation has a trig function?

jeff
Posts: 38
Joined: Mon Mar 05, 2018 11:54 pm

### Re: Given a cart and spring on incline, how do I find the period of oscillation?

Hi roxieqk16,
roxieqk16 wrote:
Mon Nov 19, 2018 6:58 pm
I don't understand your answer to the OP. For springs I can find the differential equation for a horizontal and vertical spring, but how can they be the same? The equations are

m d^2x/dt^2 + k x = 0

m d^2x/dt^2 + k x + m g = 0

How can those give the same results whenthe equation is different? And also the same when the spring is on an incline and the equation has a trig function?
The part of the motion that is the same for all three cases is the oscillatory motion. In all three cases, if the spring is at rest, but is then pulled 5 cm and released, the mass on the spring will oscillate around the "starting point", with a total length of motion of 10 cm (i.e., 5 cm away from the rest position in both sides as it goes back and forth).

There are several ways to understand that. One way is to just go ahead and solve the differential equations like you have them written. Once you see those solutions it should make sense how the motion is just shifted for each case. For example, the solution to the first one is

$x=A\sin(\omega t)+B\cos(\omega t)$

and the other is

$x=A\sin(\omega t)+B\cos(\omega t)-\frac{mg}{k}$

But I think it is better to consider the equations more directly so that we can see what is going on. Note that in your two equations, you have the spring force as kx for both cases, which means you have called the rest length zero. But the spring force depends on a change in length, or

$\mbox{spring force}= - k\ \Delta x = -k (x - x_0)$

A horizontal spring has a natural, unstretched length of L. When it is stretched a distance x, it's length is then (L+x), or

$\mbox{spring force}= -k ((L+x) - L) = -kx$

so you just get kx for the horizontal spring.

Now let's look at the vertical hanging spring. It has a natural unstretched length of L. However, when the mass is placed on it and lowered slowly, the weight stretches it an additional distance d, so that it's total length is (L+d). Note that d is that distance when the spring force and gravity forces cancel, or when

$\mbox{equilibrium}\rightarrow k((L+d)-L) - m g=0$

Finally, the mass is hanging there and then someone pulls down on the spring an additional distance and releases it. x is the deviation away from the equilibrium point, so the total length of the spring is (L+d+x) and the total force is

$\mbox{total force}=k((L+d+x)-L) - m g$

pull out a factor of kx,

$\mbox{total force}=kx+\left[ k((L+d)-L)-mg\right]$

but the meaning of d was given above, and shows that all of the that stuff in the square brackets is equal to zero, so

$\mbox{total force}=kx$

and you see we get back that the total force is just kx. (The difference in sign is just an indication of direction. Since we were pulling the spring down, the spring force acts to pull upwards.)

So that's two ways to approach this. With your two equations, the x is identical in both equations and the solutions are just shifted by a constant amount.
In the other way, the x is always the deviation away from equilibrium, so x is measured from a different point but the solutions are then identical.