Question

A grandfather clock is found to gain 39.0 s every 24 hours. If the current length of the pendulum is 0.660 m, how much should the length be changed (in m) to make the clock accurate? Let a positive value correspond to an increase in length, and a negative value correspond to a decrease.

ΔL=________

Answer 1

Step-by-step explanation:

The time period (T) of a simple pendulum depends on the length (l) of the pendulum and acceleration due to gravity (g). The time period of a simple pendulum can be calculated as follows

T = 2π√(l/g)

The number of hours in a day is 24 h.

Convert 24 h to second as follows,

1 hr = 60mins

1 min = 60secs

1 hr = 60 × 60 = 3600secs.

24hr = 24 × 3600 = 86,400sec

The number of cycle’s clock made per day is,

86,400 / 2 = 43,200 sec

If the clock runs, slow of 39 s, then there is 19.5 cycles is reduces per day. So,

(43200 - 19.5) = 43,180.5 seconds

Then, The new time period would be reduces by 43,180.5 / 43200 to the old time period

To = 43,180.5/43200 Tn

To = 0.99955 Tn.

To is old period

Tn is new period.

Given that, initial length is.

Lo = 0.66m

Relating the old period to the new period

To = 2π√(Lo / g). Equation 1

Tn = 2π√(Ln / g). Equation 2

We know that

To = 0.99955 Tn

To / Tn = 0.99955

Divide equation 1 with 2

To/Tn = 2π√(Lo / g) / 2π√(Ln / g)

To / Tn = √(Lo / Ln)

0.99955 = √(Lo / Ln)

0.99955² = Lo / Ln

0.9991 = 0.66 / Ln

Ln = 0.66 / 0.9991

Ln = 0.6606m

The difference in the lengths is,

∆L = Ln — Lo

∆L = 0.6606—0.66

∆L = 5.96 × 10^-4m

∆L = 0.596 mm

Therefore, the new length of the pendulum is 0.6606m that is the length of the pendulum should be reduced by 0.596mm.