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1 July, 07:46

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?

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  1. 1 July, 11:26
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    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
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