First slide

Applications of SHM

Question

A clock which keeps correct time at $20^{o} C$ , is subjected to $40^{o} C$ . If coefficient of linear expansion of the pendulum is $12 \times 10^{- 6} / ° C$ . How much will it gain or loose in time

Moderate

A

10.3 seconds / day
B

20.6 seconds / day
C

5 seconds / day
D

20 minutes / day

Solution

Time period $T \propto \sqrt{l} \Rightarrow \frac{ΔT}{T} = \frac{1}{2} \frac{Δl}{l} = \frac{1}{2} αΔθ$
Also according to thermal expansion $l' = (1 + αΔθ)$
$\frac{Δl}{l} = α + θ$ . Hence $\frac{ΔT}{T} = \frac{1}{2} \frac{Δl}{l} = \frac{1}{2} αΔθ$
$= \frac{1}{2} \times 12 \times 10^{- 6} \times (40 - 20) = 12 \times 10^{- 5}$
$\Rightarrow ΔT = 12 \times 10^{- 5} \times 86400$ seconds / day
$∴ ΔT \approx 10 .3$ seconds / day

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