An aluminum clock pendulum having a period of 1.00 s keeps perfect time at 20.0°c.
(a) when placed in a room at a temperature of-4°c, will it gain time or losetime?
it will lose time.
it will gain time.
(b) how much time will it gain or lose every hour?

a) T ’= 0.999 s ,  b)  t = 3596.4 s

Explanation:

The angular velocity of a simple pendulum is

        w = √g / L

The angular velocity, frequency and period are related

        w = 2π f = 2π / T

        2π / T = √ g / L

        T = 2π √ L / g

        L = T² g / 4π²

        L = 1² 9.8 / 4π²

        L = 0.248 m

To know the effect of the temperature change let's use the thermal expansion ratios

       ΔL = α L ΔT

       ΔL = 24 10⁻⁶ 0.248 (-4 - 20)

       ΔL = 142.8 10⁻⁶ m

       Lf - L = -142. 8 10⁻⁶

       Lf = 142.8 10⁻⁶ + 0.248

       Lf = 0.2479 m

Let's calculate new period

      T ’= 2π √ L / g

      T ’= 2π √ (0.2479 / 9.8)

      T ’= 0.999 s

We can see that the value of the period is reduced so that the clock is delayed

b) change of time in 1 hour

When the clock is at 20 ° C in one hour it performs 3600 oscillations, for the new period the time of this number of oscillations is

       t = 3600 0.999

       t = 3596.4 s

Therefore the clock is delayed almost 4 s