The time represented by the clock hands of a pendulum clock depends on the number of oscillations performed by pendulum. Every time it reaches to its exterme position the second hand of the clock advances by one second that means second hand moves by two second when one oscillation is completed.
(a) How many number of oscillations completed by pendulum of clock in `15` minutes at calibrated temperature `20^(@)C`
(b) How many number of oscillations are completed by a pendulum of clock in `15` minutes at temperature of `40^(@)C if alpha = 2 xx 10^(-5)//"^(@)C`
(c ) What time is represented by the pendulum clock at `40^(@)C` after `15` minutes if the initial time shown by the clock is `12:00` pm?
(d) If the clock gains two seconds in `15` minutes in correct clock then find-(i) Number of extra oscillations (ii) New time period (iii) change in temperature.

Answer

(a) `450` (b) `449`
(c ) `12:14:59 pm`
(d) (i) `1` (ii) `(900)/(450)s` (iii) `(10^(5))/(450)"^(@)C`

Explanation

(a) Number of oscillations `= (15 xx 60)/(2) = 450`
(b) `DeltaT = (1)/(2) xx Delta theta T = (1)/(2) xx2xx 10^(-5) xx20 xx2 = 4 xx 10^(-4) sec`.
`T' = 2 +4 xx 10^(-4) = 2.004 sec`.
Number of oscillations `= (15 xx 60)/(2.0004) = 449.91`
Number of complete oscillations `= 449`
(c ) Last oscillation will not be complete.
So, time shown `= 12:15 :00 - 0:00:01 = 12:14:59 pm`
(d) `2 xx (1)/(2) x2xx 10^(-5) xx Delta theta xx 15 xx 60`
`rArr Delta theta = (2 xx 10^(5))/(15 xx 60) = (10^(5))/(450)"^(@)C`
`DeltaT = (1)/(2) alpha Delta theta T = (1)/(2) xx2xx 10^(-5) xx (10^(5))/(450) xx2 = (2)/(450)`
`T' = 2 - (2)/(450) = (898)/(450) sec`.
Number of oscillations `= (15 xx 60 xx 450)/(898) = 451`
Number of extra oscillations `= 451 - 450 = 1`.