The period T of a simple pendulum is the time taken for one complete small oscillation. T varies directly with the square root of its length L. When the length of the pendulum is 98 cm, the period is exactly 2 seconds.

i) What length should the pendulum be if its period is to be halved?

ii) A grandfather clock is regulated by this pendulum. The pendulum is lengthened to 1 m and the clock started at noon. Calculate the actual time, to the nearest second, when the clock first shows 1 o'clock.

For i) I worked out the length as 1/4 and ii) the time is 2:01. My answers dont look right. Could someone please help?

2. Originally Posted by xwrathbringerx
The period T of a simple pendulum is the time taken for one complete small oscillation. T varies directly with the square root of its length L. When the length of the pendulum is 98 cm, the period is exactly 2 seconds.

i) What length should the pendulum be if its period is to be halved?

ii) A grandfather clock is regulated by this pendulum. The pendulum is lengthened to 1 m and the clock started at noon. Calculate the actual time, to the nearest second, when the clock first shows 1 o'clock.

For i) I worked out the length as 1/4 and ii) the time is 2:01. My answers dont look right. Could someone please help?

As the period varies directly with the square root of its length you have:

T=k\sqrt{L}

for some constant k.

With L in metres, you are told that:

$\displaystyle T=k\sqrt{0.98}=2$ ,

so:

$\displaystyle k=2/\sqrt{98}$

Now what is the period when $\displaystyle L=1$?

One hour is 1800 swings of a 2 second pendulum, at what time does the 1 metre pendulum complete 1800 swings?

RonL

3. Ummm is it 37 s past 1 o'clock?

4. Originally Posted by xwrathbringerx
Ummm is it 37 s past 1 o'clock?
The period of the $\displaystyle 1$ m pendulum is:

$\displaystyle \tau=2/\sqrt{0.98}$ s

The time for $\displaystyle 1800$ swings is:

$\displaystyle 1800 \times \tau=3600/\sqrt{0.98}$ s

which is:

$\displaystyle 1800 \times \tau-2\times 1800=3600/\sqrt{0.98}-3600=36.55$ s

longer than $\displaystyle 1800$ swings of the $\displaystyle 2$ s pendulum.

So yes, the time rounded to the nearest second is $\displaystyle 37$ s past $\displaystyle 1$ o'clock

RonL