Question

if the length of second pendulumn is increased by 1% how many seconds will be lost in day

Answer 1

Explanation:

The time period of a second pendulum is 2s.

Now, the time period of a pendulum varies directly as the square root of its length i.e.:

T∝l√(1)

From (1):

T′T=l′l−−√(2)

l′=l+2 % of l=1.02l(2.5)

∴l′l=1.02(3)

Putting (3) in (2):

T′T=1.02−−−−√≈1.01(4)

T′=1.01T=2.02 s(5)

Now, the clock will complete one oscillation in 2.02 s but will show the time per oscillation as 2 s.

Therefore, the clock will lose 0.02 s for every 2 s⟹ it will lose 0.01 s for every 1 s.

Thus, in one day( 86400 s) it will lose 0.01⋅86400=864 s.

In equation (4), I used the binomial approximation for x≪1,

(1+x)a≈1+xa

The actual result is about

1.009504938...

Which makes the actual time lost about

859.72266432...