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A very practical question. So, let's imagine that we have a small cart attached to a spring lying on an inclined plane. If I move my cart by 10 cm (measured as if there were a ruler lying along the inclined plane) such that the spring is compressed, and then release it, will my cart oscillate in simple harmonic motion with an amplitude of 10cm measured along the plane?

I thought about this for some time now. But, I think the answer is simply no. Because there will always be the impact of a component of the force of gravity. So, the cart should still move in SHM, but not with an amplitude of 10 cm, right? If I am wrong, please explain with detail.

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  • $\begingroup$ The setup of this lab is very similar to p. 21 of ocw.mit.edu/courses/physics/… The amplitude of the best fit is obtained from the data obtained by an ultrasound detector at the bottom of the inclined plane. My lab manual says that there will be a difference between the amplitude set and the amplitude of the fit. $\endgroup$ – yolo123 Aug 26 '14 at 1:14
  • $\begingroup$ Do you mean SPRING? If you think the amplitude is less can you explain where the difference in gravitational potential energy has gone? This looks like a homework-type question to me. $\endgroup$ – Rob Jeffries Aug 26 '14 at 16:41
  • $\begingroup$ Oh boy, do pardon the fact that English is my 2nd language. $\endgroup$ – yolo123 Aug 27 '14 at 1:41
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The force of gravity will simply move the equilibrium point, being a costant force. So if $l_0$ is the initial lenght of the spring, the applying the orizontal component of the force $Mg\sin \theta$, you'll get the new lenght in equilibrium that is (Hooke's Law) $$ \vec{F} = -k\vec{(x - x_0)} \Rightarrow x = F/k + x_0$$ $$x_1 = Mg\sin \theta / k + x_0$$ Since gravity is costant over time, any displacement applied to the string will act as if the initial lenght of the string is $x_1$, not $x_0$. So the amplitude of harmonic oscillation won't change, but you must consider the amplitude relative to $x_1$, not $x_0$. You have to put the $0$ in your ruler on $x_1$, rather than $x_0$.

You can prove this statement solving the differential equation: $$F = ma = m\frac{d^2x}{dt^2} = -k(x - x_0) + Mg\sin \theta $$ $$ m\frac{d^2x}{dt^2} +kx = kx_0 + Mg\sin \theta $$ there is no $x$ in the $Mg\sin \theta$ so you can simply consider $$kx_0 + Mg\sin \theta$$ as a costant term $kx_1$ so you'll get $$ \frac{d^2x}{dt^2} + \frac{k}{m}x = \frac{k}{m}x_1 $$ the amplitude is given solving the differential equation http://en.wikipedia.org/wiki/Ordinary_differential_equation it doesn't depend on $x_0$ or $x_1$.

Considering the problem using conservation of energy, you'll get: Let's say that $h_0$ is the height relative to $x_0$ so $h = x\cos \theta $ energy at $x_0$ will be $$ mgh_0 + \frac{1}{2}m\upsilon^2 + \frac{1}{2}k(x_0 - x_0)^2$$ energy at a displacement $\Delta x$ from $x_0$ $$mgh_0 + mg\Delta h + \frac{1}{2}m\upsilon^2 + \frac{1}{2}k(\Delta x)^2$$ every force is conservative so $$ mgh_0 + \frac{1}{2}m\upsilon_0^2 + \frac{1}{2}k(x_0 - x_0)^2 = mgh_0 + mg\Delta h + \frac{1}{2}m\upsilon_1^2 + \frac{1}{2}k(\Delta x)^2$$ if the cart is release with $\upsilon_1 = 0$ $$ \frac{1}{2}m\upsilon_0^2 = mg\Delta x \cos \theta + \frac{1}{2}k(\Delta x)^2$$ the speed at $x_0$ depends only from the inital displacement $\Delta x$ and not from $x_0$ or $x_1$ absolutely.

Theoretically the amplitude won't change, but since there are other factors as friction, measures aren't perfectly matching to theory. The inclined plane present less friction rather than an horizontal plane due to the vertical component of the weight of the cart. So taking friction into account total energy will decrease while the cart is moving so, it will decrease the amplitude of motion.

Hope this help!

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  • $\begingroup$ Perfect. That is what I thought. So detailed. So well-explained. My lab protocol wasn't really clear on what they meant. But now I get it. $\endgroup$ – yolo123 Aug 27 '14 at 1:43

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