Question

A torsional pendulum consists of a solid disc connected to a thin wire (α = 2.4 × 10–5 °C–1) at its centre. Find the percentage change in the time period between peak winter (5°C) and peak summer (45°C).

Answer 1

The percentage change in the time period between peak winter (5°C) and peak summer (45°C) is $9.6 \times$ $10^{-2}$ percent

Explanation:

Step 1:

The wire  Coefficient of linear expansion, $\alpha = 2.4 \times 10^{-5}$   $^{\circ}C^{-1}$

Let $I_0$  be the torsional pendulum's moment of inertia at 0 ° C

If K is the wire's torsional constant, then Torsional pendulum time period T:

$T=2\pi \sqrt{\frac{I}{K}$………….(1)  

Here

I = Moment of inertia a after temperature change

When the temperature changes by ∆θ, moment of inertia I,

$I = I_0(1+2\alpha \Delta \theta)$

Step 2:

When the value of I is replaced in equation(1) we get:

$T_{1}=2 \pi \sqrt{\frac{I_{o}((1+2 \alpha \Delta \theta))}{K}}$

∆θ= 5 °C

Time period $T_1$

$T_{1} &=2 \pi \sqrt{\frac{l_{0}((1+2 \times \alpha \times 5))}{K}}$

$T_{1} &=2 \pi \sqrt{\frac{l_{0}((1+10 \alpha))}{K}}$

∆θ= 45 °C

Step 3:  

Time period $T_2$

$T_{2}=& 2 \pi \sqrt{\frac{l_{0}(1+2 \times \alpha \times 45)}{K}}$

$& T_{2}=2 \pi \sqrt{\frac{l_{0}(1+90 \alpha)}{K}}$

$\frac{T_{2}}{T_{1}}=& \frac{2 \pi \sqrt{\frac{l_{0}(1+90 \alpha)}{K}}}{2 \pi \sqrt{\frac{\left.l_{0}(1+10 \alpha)\right)}{K}}}$

$& \frac{T_{2}}{T_{1}}=\sqrt{\frac{1+90 \alpha}{1+10 \alpha}}$

$\frac{T_{2}}{T_{1}}=& \sqrt{\frac{1+90 \times 2.4 \times 10^{-5}}{1+10 \times 2.4 \times 10^{-5}}}$

$\frac{T_{2}}{T_{1}}=\sqrt{\frac{1.00216}{1.0024}}$

$\left.\%_{ \text { change }}=\left( \frac{T_{2}}{T_1}\right.}-1\right) \times 100$

$&=(\sqrt{\frac{1.00216}{1.0024}}-1) \times 100$

Consequently, the percentage change in the torsional pendulum period between peak winters and peak summers is 9.6 x  $10^{-2}$ percent

$&=0.0959 \%=9.6 \times 10^{-2} \%$

Answer 2

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The percentage change in the time period between peak winter and peak summer is

$9.6 \times {10}^{ - 2}$