A copper pendulum of period T 0 on the ground is taken into a mine shaft to depth d . Its expansion coefficient is α , and the mine is θ ∘ C warmer than the ground. Its period becomes T . Find T 0 T to first order. R is the Earth radius; ignore density variations in the crust.

L ′ = L ( 1 + α θ ) and g ′ = g ( 1 − R d ) T 0 T = 1 − R d 1 + α θ To first order, T 0 T ≈ ( 1 + 2 α θ ) ( 1 + 2 R d )

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A copper pendulum of period $T_{0}$ on the ground is taken into a mine shaft to depth $d$ . Its expansion coefficient is $α$ , and the mine is $θ^{\circ} C$ warmer than the ground. Its period becomes $T$ . Find $\frac{T}{T _{0}}$ to first order. $R$ is the Earth radius; ignore density variations in the crust.

Options:

Answer:

Solution:

$L^{'} = L (1 + α θ)$
and
$g^{'} = g (1 - \frac{d}{R})$
$\frac{T}{T _{0}} = \frac{1 + α θ}{1 - \frac{d}{R}}$
To first order,
$\frac{T}{T _{0}} \approx (1 + \frac{α θ}{2}) (1 + \frac{d}{2 R})$

Stream:JEESubject:PhysicsTopic:Heat & ThermoSubtopic:Thermal Expansion

⏱ 2mℹ️ Source: QB

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