Two side-by-side pendulum clocks have heavy bobs at the ends of rigid, very ligh

Question

Two side-by-side pendulum clocks have heavy bobs at the ends of rigid, very lightweight arms. One pendulum has a 38.8 cm -long rod, the other a 24.8 cm -long rod. Each clock makes one tick for each complete swing of its pendulum.

Determine the frequency of the first pendulum.

Determine the period of the first pendulum.

Determine the frequency of the second pendulum.

Determine the peirod of the second pendulum.

Because the two pendulums have different frequencies, their ticks are usually "out of step." However, you notice that they do get back into step (tick at the same instant) at regular intervals. How much time elapses between such events?

The getting-into-step phenomenon is, itself, periodic. What is the frequency of this phenomenon?

Explanation / Answer

Solution: From the question we have

Two side-by-side pendulum clocks have heavy bobs at the ends of rigid very light.

One pendulum has a 38.8 cm -long rod, the other a 24.8 cm -long rod

Each clock makes one tick for each complete swing of its pendulum.

Now,

The period os a simple pendulum is 2*sqrt(L/g) where g = acceleration due to gravity 980cm/s^2

So the first pendulum has a period (T) of 2*sqrt(38.8cm/980cm/s^2) = 1.25s
Therefore the frequency f = 1/T = 1/1.25s = 0.800Hz

So the second pendulum has a period (T) of 2*sqrt(24.8cm/980cm/s^2) = 1.00s
Therefore the frequency f = 1/T = 1/1.25s = 1.00Hz

The two pendulum are in step when the number of cycles of the first n1*1.25 is equal to the number of cycles of the second n2*1.00
So 1.25n1 = 1n2

So when n1 = 4 n2 = 5..so every 5.0s these are in step.

So if T = 5.0s, then the f = 1/5 = 0.20Hz

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.