# Simple harmonic motion (with calculus)

In summary, Adrian attempted to solve the homework equation by finding the trajectory and the maximum force acting on the object at any given time. However, he ran into some difficulty with the dimensional inconsistency in the expression 8b-4ab. He needs help from the tutor to figure out where the inconsistency arises.

## Homework Statement

An object with mass m undergoes simple harmonic motion, following 2 perpendicular directions, described by the equations:

x=a cos (wt), a>0,
y=b cos (2wt), b>0

a) find the equation of the trajectory
b) find the speed at any given time (so having t as a variable)
c) the maximum force F which acts on the object at any given time (again, having t as a variable).

## The Attempt at a Solution

So far: a) from x=a cos (wt) we get cos (wt)=x/a; in the y equation, we can expand as follows:
cos (2wt)=cos^2 (wt) - sin^2 (wt). We also know that for any real x we have cos^2 (x) +sin^2(x)=1, therefore cos (2wt)=cos^2 (wt) - sin^2 (wt)=cos^2 (wt) + cos^2(wt)-1; therefore y=b (2cos^2 (wt) -1 )=b( x^2/a^2 -1 ), which is the equation of the trajectory.

Now for b) and c), I'm not quite sure how to use what I have. I differentiated the x and y equation from the beginning, differentiated the trajectory and somehow I need to combine them. I suppose the idea from b) applies to c).

I would be grateful if you could give me some hints :)

Have a great day,
PS: If it is not clear, I could rewrite using latex.

Solved b), only need further help at c) (got that Fmax=mw^2 (8b-4ab)/a , which doesn't seem correct to me )

Solved b), only need further help at c) (got that Fmax=mw^2 (8b-4ab)/a , which doesn't seem correct to me )
8b-4ab is dimensionally inconsistent, so cannot be right. Step through your working to find where the dimensional inconsistency arises.

Apparently, I forgot that since those 2 directions are perpendicular, a= sqrt ( ax^2+ay^2) , where ax is the acceleration found by differentiating vx from b) and the same for ay. Basically, I have to maximize sqrt ( a^2*w^4*cos^2 (wt) + 16*b^2*w^4*cos^2( 2wt) ), which doesn't require calculus, the maximum is obviously achieved when both cosines are =1, and therefore the maximum acceleration would be sqrt (a^2*w^4+16*b^2*w^4). I might be missing sth though, does that seem correct?

Apparently, I forgot that since those 2 directions are perpendicular, a= sqrt ( ax^2+ay^2) , where ax is the acceleration found by differentiating vx from b) and the same for ay. Basically, I have to maximize sqrt ( a^2*w^4*cos^2 (wt) + 16*b^2*w^4*cos^2( 2wt) ), which doesn't require calculus, the maximum is obviously achieved when both cosines are =1, and therefore the maximum acceleration would be sqrt (a^2*w^4+16*b^2*w^4). I might be missing sth though, does that seem correct?
There is one step missing in the reasoning above, though perhaps you merely omitted to mention it: that both cosines can be 1 simultaneously.

Not sure if I understand your point, by saying "the maximum is obviously achieved when both cosines are =1" I meant to say that the maximum occurs when both cosines are equal to 1 (when t=2k pi, k integer). Is this correct, sir?

when t=2k pi, k integer
Yes, that's the part you needed to add for completeness. E.g. if the expression had been ##\cos(\omega t)+\cos(\omega t+\cos(t))## it might not be possible for that ever to equal 2.

I see your point, thank you for taking your time to help me, haruspex. Wish you an amazing day!

## 1. What is simple harmonic motion?

Simple harmonic motion is a type of periodic motion in which an object oscillates back and forth around an equilibrium point, with a restoring force that is directly proportional to the displacement from the equilibrium point. This type of motion can be seen in systems such as a swinging pendulum or a mass-spring system.

## 2. How is simple harmonic motion described in terms of calculus?

Simple harmonic motion can be described using the equation x(t) = A sin(ωt + φ), where x is the displacement from equilibrium, A is the amplitude, ω is the angular frequency, and φ is the phase constant. This equation can be derived using calculus by solving the second-order differential equation for the motion.

## 3. Can you provide an example of simple harmonic motion using calculus?

One example of simple harmonic motion using calculus is a mass-spring system. When a mass is attached to a spring and pulled down, it will oscillate up and down around the equilibrium point at a frequency determined by the mass and the spring constant. This motion can be described using the simple harmonic motion equation and can be analyzed using calculus.

## 4. What are some real-world applications of simple harmonic motion?

Simple harmonic motion has many real-world applications, including pendulum clocks, musical instruments, and shock absorbers in cars. It is also used in engineering to design buildings and bridges that can withstand vibrations caused by wind or earthquakes.

## 5. How does the amplitude affect simple harmonic motion?

The amplitude of simple harmonic motion is the maximum displacement from equilibrium. A larger amplitude will result in a greater distance traveled during each oscillation, while a smaller amplitude will result in a shorter distance. This also affects the energy of the system, with a larger amplitude resulting in a greater amount of energy exchanged between potential and kinetic energy during each oscillation.

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