Your grandfather clock's pendulum has a length of 0.9930 m. Part A If the clock runs slow and loses 19 s per day, how should you adjust the length of the pendulum? Note: due to the precise nature of this problem you must treat the constant g as unknown (that is, do not assume it is equal to exactly 9.80 m/s2).

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

T= 2[tex]\pi \sqrt{\frac{l}{g}[/tex]

The number of hours in a day is 24 h.

Convert 24 h to second as follows,

24 x 60 x 60 =86400 s

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

86400/2 = 43200

If the clock runs, slow of 19 s, then there is 9.5 cycles is reduces per day. So,

( 43200 -9.5) cycles = 43190.5 cycles

The period would be reduces by. 43190.5/43200 = 0.9998

Then the new time period is,

T new = 0.9998 Told

Substitute

[tex]\sqrt{LNEW} =0.9998\sqrt{LODD}[/tex]

LNEW/LODD =[tex] 0.9998^{2}[/tex]

LNEW/LODD =0.9996

Substitute 0.9930 m for LODD

LNEW = 0.9930 X 0.9996 = 0.9926

The difference in the lengths is,

LNEW- LODD

= 0.9926- 0.9930 = -0.0004

This is the same as 0.4mm

Therefore, the new length of the pendulum is 0.9926m, the length of the pendulum should be reduced by. 0.4mm

okpalawalter8 okpalawalter8 · Answer 2 · 2022-04-12T12:14:02+02:00

The pendulum's adjustment is mathematically given as

dL=0.4mm

What adjustment should be done to the pendulum?

Question Parameter(s):

Your grandfather clock's pendulum has a length of 0.9930 m

If the clock runs slow and loses 19 s per day

Generally, the equation for the period is mathematically given as

[tex]T= 2\pi \sqrt{\frac{l}{g}[/tex]

Where, total cycles is

Tc=( 43200 -9.5) cycles

Tc = 43190.5 cycles

Therefore

T new = 0.9998 Told

Lnew/Lold =0.9998^{2}

Hence

Lnew = 0.9930 X 0.9996

Lnew= 0.9926

In conclusion, difference in the lengths as

dL = 0.9926- 0.9930

dL = -0.0004

dL=0.4mm