Question:

If we use pendulum to measure gravity. The length is known exactly but stopwatch has least count of 1 second, minimum error will be if there are?

• 10 oscillation in 40 seconds
• 20 oscillation in 60 seconds
• 20 oscillation in 40 seconds
• 30 oscillation in 60 seconds

Solution

We know that time period of a pendulum can be calculated using T=2pisqrt(l/g).

Now from this we get that (deltaT)/T = 1/2 (deltag)/g.

To find (deltag)/g we need the time period of the pendulum and the error in measuring time period.

Do you think the least count of stop watch is the error in time period? No it not.

Let us share our reasoning.

Time period T = t/n where n is the number of oscillations and t is the time taken for n oscillations.

So (deltaT)/T = (deltat)/t

=> deltaT= t/nxx(deltat)/t

=> deltaT =1/n deltat=1/n " since " deltat=1

So (deltag)/g = 2(deltaT)/T = 2xx((1/n) / (t/n)) = 2/t.

The error in calculating gravity will be less for the pendulum which is observed for a longer duration.