0
$\begingroup$

We have the following linear 2nd order differential equation that describes its harmonic motion of a pendulum spring:

$$ y'' + \frac{k\cdot y}{m} = 0 $$

where $m$ is the mass of the pendulum ball, $k$ is the proportionality factor from Hook's law, and $y$ is the elongation from equilibrium position. A possible solution is

$$ y(t) = y_{max} \sin{(\frac{2 \pi}{T}t + \rho}) $$

How to derive an expression for the maximum speed?

Attempt:

I start by differentiating $y(t)$ to get $v(t)$:

$$ v(t) = \frac{2\pi}{T}y_{max} \cos\left( \frac{2\pi}{T}t + \rho \right) $$

Then I optimize $v(t)$ by settings its derivative equal to zero: $\frac{dv(t)}{dt} = 0$

$$ \frac{dv(t)}{dt} = - \frac{4 \pi^2}{T^2}y_{max} \sin \left( \frac{2\pi}{T}t + \rho \right) = 0 $$

However, I am not sure how to go on from here to actually obtain an expression for the maximum speed. Intuitively I understand that the speed is highest when the pendulum reaches the equilibrium position, when $y=0$, or equivalently when $a=0$ - of course, the acceleration is zero at $y_{max}$ also, so perhaps I should use the two conditions that both $a=y=0$, somehow?

$\endgroup$
2
$\begingroup$

The maximum speed is just the amplitude of $v\left(t\right)$, i.e. $\dfrac{2\,\pi}{T}\,y_{\text{max}}$, and you will get it the first time if the argument of the cosine equals to 0.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.