Find the time interval between oscillations of SHM.

Question

Parts i) and ii) I can solve.

But for part iii) I can't do, as I don't know which equation describes the SHM motion? Is it $y=0.5\sin(1.2t)$ or $y=0.5\cos(1.2t)$ or $x=0.5\sin(1.2t)+2.5$?

I thought the correct approach was to differentiate $x=0.5\sin(1.2t)+2.5$ with respect to time $t$ and equate to $0.48$ and solve to get the $2$ different times. But it isn't working.

Could someone please help me solve part iii)?

Here is the official answer:

What I would like to know is how this "$\left[t_0 = \cfrac{\sin^{-1}(0.6)}{1.2};\space t_1 = \cfrac{\cos^{-1}(0.6)}{1.2}\right]$" was deduced?

With best regards.

Answer 1

Your third equation for SHM should completely describe the system up to some (indeterminable) phase shift. Your method of differentiating the equation should work fine for part iii. Did you consider that "speed" is not the same as "velocity"? ie, set |X'|=0.48, and not X'=0.48. Also, since P is moving "towards" O at $t_1, t_2$, it makes sense to use the cosine function: $$\Big|\frac{d(0.5 cos(1.2 t)+2.5)}{dt}\Big|=0.48$$

$$t = \frac{5}{6} \Big(2 \pi n \pm sin^{-1}\frac{4}{5}\Big),\ \ \ \ n\in\mathbb{Z} $$ $$t = \frac{5}{6} \Big(2 \pi n +\pi \pm sin^{-1}\frac{4}{5}\Big),\ \ \ \ n\in\mathbb{Z} $$ $$t_1=0.7727,\ t_2=1.8452,\ t_3=3.3907$$ $$\Delta t_{12}=1.0725,\ \Delta t_{23}=1.5455$$

Answer 2

All three of your equations describe the same SHM.

In part (iii), we are interested only in velocity and time, so use $\frac{dx}{dt}=(0.5)(1.2) \cos(1.2t) $ Consider the cos graph from $0$ to $2\pi$, which is $0$ to 5.4 in our problem (sorry I dont know how to draw it here). Consider $\vert \frac{dx}{dt} \vert = 0.48 $. On the graph, where the gradient is negative, this occurs where $ \frac{dx}{dt} = 0.48 $ and $ \frac{dx}{dt} = -0.48 $. Solving these equations gives $t_1 = 0.54$, and $ t_2 = 2.08$. And from the symmetry of the graph, the next two values (where the gradient is positive) are $t_3 = 5.24 - 2.08$ and $t_4 = 5.24 - 0.54$.