0
$\begingroup$

I am taking a first year undergrad physics class (waves & modern physics), and the answer my book gives to the question bellow has left me a bit confused. Here is the question (excuse me if it’s not the best translation from my book that is written in French):

A pendulum that consists of a string of 2m length and a 10kg cube attached to its extremity is immobile. A 0.01 kg bullet is fired at 400 m/s, and is eventually incrusted inside the cube.
Supposing the time it takes for the bullet to incrust the cube is insignificant, calculate the interval of time it takes between the impact of the bullet and the first instant where the pendulum is at the highest point of its trajectory.

The book’s answer key simply states that upon calculation of the angular frequency, given by $\omega_{0} = \sqrt{\frac{g}{L}}$, and then the period T, given by $T = \frac{2\pi}{\omega_{0}}$, I can just divide the period by 4 ($\Delta t = \frac{T}{4}$) and that will give me the time of the highest point of the trajectory.

I cannot seem to understand why taking the fourth of the period would give the highest point of the pendulum.

$\endgroup$
1
$\begingroup$

As you can see from the formula for the period $$T=2\pi\sqrt{\frac{L}{g}}$$ it does not depend on the energy or the amplitude of the pendulum. Independent of how hard you push it, the frequency will always be the same. Be very careful though, this statement only holds for "small" energies, because then $\sin(\theta) \approx \theta$. Otherwise it will start to deviate from a harmonic motion.

Anyway, since in this approximation the energy does not influence the period, we can ignore the specifics of the bullet and say the highest point is reached after $T/4$, because $T$ is the time it takes for the pendulum to go from the center to the left, come back down to the center, go over to the right, and finally come back down to the center again (at which point everything is the same as in the beginning). $T$ is precisely defined in this way; it is the time it takes for the pendulum to reach its initial state again. Since all 4 of these parts take the same amount of time, only one of these parts (going from the center all the way up to its highest point) takes $T/4$.

$\endgroup$
2
  • $\begingroup$ I keep forgetting to visualize the mouvement of pendulums/springs as sine/cosine functions. You’ve made it all clear about the fourth of the period being the maximum of the amplitude reached, since at t=0, x was also at 0. Thank you so much! $\endgroup$ Feb 26 '19 at 0:25
  • 2
    $\begingroup$ It might be noted that the equipartition of time among the four stages of motion stays true simply via the symmetry of the problem (and the assumption that the system is without friction etc.) and doesn't require the small-amplitude approximation. $\endgroup$
    – Dvij D.C.
    Feb 26 '19 at 0:30
0
$\begingroup$

The time period of the pendulum $T$ which relates to its angular frequency $\omega$ via $T=\frac{2\pi}{\omega}$ is defined as the time it takes for the pendulum to complete a full swing (i.e., (i) from its minimum height to its maximum height on one side (ii) to its minimum height (iii) to its maximum height on the other side (iv) and finally, to its minimum height again). In an ideal pendulum with no friction, simply via the symmetry of the problem, all these four stages of motion take an equal amount of time--and thus, any one stage of them takes a quarter of the total time period.

The crucial point to note here is that the time period of the pendulum with small amplitudes is not dependent on the energy of the pendulum--it only depends on the length of the pendulum (and the gravitational acceleration). Thus, the simple fact the pendulum is set in motion automatically tells us that it will take $T$ amount of time to complete a full swing regardless of its energy.

$\endgroup$

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.