# Question e1d2a

Sep 10, 2016

We are not given the radius of the pendulum, so we cannot calculate its change in volume due to the new temperature.

Besides that, even if we could calculate the change in volume, its mass does not change while its density does, which does not affect the force of gravity perpendicular to the pendulum bob and thus does not affect its period.

When we look at an ideal pendulum, we have that it has length $L$, and it is a non-massless pendulum of mass $m$ with a massless string. Note that the magnitude of the tension is equal to the parallel component of the force due to gravity. So, the sum of all forces is:

$\sum {\vec{F}}_{y} = \vec{T} - m \vec{g} \cos \theta = 0 \implies \vec{T} = m \vec{g} \cos \theta$
$\sum {\vec{F}}_{x} = m g \sin \theta \ne 0$

The arc length of travel is $s = L \theta$, so the restoring force is analogously ${F}_{\text{res}} = - k x = - k L \theta$, where $k$ is the force constant.

Note that the restoring force is opposite in direction to the component of the force due to gravity perpendicular to the pendulum.

So, $| {F}_{\text{res}} | = k L \theta = m g \sin \theta$

$k = \frac{m g \sin \theta}{L \theta}$

For small $\theta$, we assume that $\sin \theta \approx \theta$ since ${\lim}_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$, i.e. $\sin \theta \to \theta$ as $\theta \to 0$. So...

$k \approx \frac{m g}{L}$

The period is defined as $\frac{2 \pi}{T} = \omega$, where $\omega$ is the angular frequency, so $T = \frac{2 \pi}{\omega}$. With $\omega = \sqrt{\frac{k}{m}}$, we have:

$\textcolor{b l u e}{T} = 2 \pi \sqrt{\frac{m}{k}}$

$= 2 \pi \sqrt{\frac{m}{\frac{m g}{L}}} = 2 \pi \sqrt{m \cdot \frac{L}{m g}} \approx \textcolor{b l u e}{2 \pi \sqrt{\frac{L}{g}}}$

This means the period of an ideal pendulum does not depend on the mass of a pendulum that has a presumably massless string.

I assume you have $\alpha = 3.6 \times {10}^{- 5} {\text{^@ "C}}^{- 1}$ for your isobaric expansion coefficient. $\alpha$ is defined as:

$\alpha = \frac{1}{V} {\left(\frac{\partial V}{\partial T}\right)}_{P}$

This is saying that $\alpha$ is a constant-pressure process in which the volume of the substance changes due to the change in temperature. $\alpha$ is valid for any substance, not just gases.

Note that since we found that the mass of the pendulum bob doesn't make a difference in its period, its period does not change when its volume changes either.

We needed the radius of the pendulum bob, and we also needed to know whether the string was considered massless or not.

Sep 12, 2016

## Laws of Simple Pendulum are

1st law or the law of isochronism:

The time period (T) is constant, when effective length (L) and acceleration due to gravity (g) are constants. This means that a pendulum will take same time in completing each oscillation, whatever is the amplitude, provided the latter does not exceed ${4}^{\circ}$. So the time period of oscillation of a simple pendulum is independent of amplitude of vibration, provided its amplitude does not exceed ${4}^{\circ}$.

2nd law or the law of length:

When acceleration due to gravity (g) is constant, the time period (T) of oscillation of a simple pendulum is directly proportional to the square root of its effective length (L).

So, $T \propto \sqrt{L}$, when g is constant.

3rd law or the law of acceleration:

When effective length (L) is constant, the time period (T) of oscillation of a simple pendulum is inversely proportional to the square root of the acceleration due to gravity (g) at a place of observation.

So, $T \propto \sqrt{\frac{1}{g}}$ when L is constant.

4th law or the law of mass:

The time period of oscillation of a pendulum at a place independent of mass and material of the bob provided the effective length of the pendulum is constant.

The equation of simple pendulum can be shown to be

$T = 2 \pi \sqrt{\frac{L}{g}}$

According to the 2nd law $T \propto \sqrt{L}$, when g is constant.

With increase in temperature there occurs thermal expansion of effective length of the solid pendulum rod causing enhancement of the Time period of the pendulum. This means the clock slows down with time as temperature increases .

To establish the relation between linear ($\alpha$ )and cubical ($\gamma$) expansion coefficient of the material of the solid pendulum rod let us consider a cube of length $l$. So its volume $V = {l}^{3}$

Now by definition linear expansion coefficient $\alpha = \frac{1}{l} \frac{\mathrm{dl}}{d \theta}$,where $\theta$ representing temperature

And cubical expansion coefficient $\gamma = \frac{1}{V} \frac{\mathrm{dV}}{d \theta}$

$\implies \gamma = \frac{1}{V} \frac{\mathrm{dV}}{d \theta} = \frac{1}{l} ^ 3 \frac{d \left({l}^{3}\right)}{d \theta} = \frac{3}{l} \frac{\mathrm{dl}}{d \theta} = 3 \alpha$

Given

$\text{Coefficient of cubical expansion of iron}$
$\left({\gamma}_{F e}\right) = 36 \times {10}^{-} 6 {C}^{-} 1$

$\text{Coefficient of linear expansion of iron}$
$\left({\alpha}_{F e}\right) = \frac{{\gamma}_{F e}}{3} = \frac{36}{3} \times {10}^{-} 6 {C}^{-} 1 = 12 \times {10}^{-} 6 {C}^{-} 1$

It is also given that pendulum keeps correct time at ${20}^{\circ} C$.

Let the effective length of the pendulum be ${L}_{0}$ at ${20}^{\circ} C$. and ${L}_{\theta}$ at $\Delta \theta$ rise of temperature and the corresponding Time periods are ${T}_{0} \mathmr{and} {T}_{\theta}$ respectively.

So ${L}_{\theta} = {L}_{0} \left(1 + {\alpha}_{F e} \times \Delta \theta\right)$

Now by 2nd Law of simple pendulum

${T}_{\theta} / {T}_{0} = \sqrt{{L}_{\theta} / {L}_{0}} = \sqrt{\frac{{L}_{0} \left(1 + {\alpha}_{F e} \times \Delta \theta\right)}{L} _ 0} = {\left(1 + {\alpha}_{F e} \times \Delta \theta\right)}^{\frac{1}{2}}$

$\implies {T}_{\theta} / {T}_{0} = \left(1 + \frac{1}{2} \times {\alpha}_{F e} \times \Delta \theta\right)$,neglecting higher power of $\alpha$

$\implies \frac{{T}_{\theta} - {T}_{0}}{T} _ 0 = \left(\frac{1}{2} \times {\alpha}_{F e} \times \Delta \theta\right)$

So the Variation in time per sec is given by this relation .

In one day $= 24 h r = 24 \times 60 \times 60 s = 86400 s$ the total variation( lose ) will be given by

$\implies \frac{{T}_{\theta} - {T}_{0}}{T} _ 0 \times 86400 = \left(\frac{1}{2} \times {\alpha}_{F e} \times \Delta \theta\right) \times 86400 s$

$= \frac{1}{2} \times 12 \times {10}^{-} 6 \times 20 \times 86400 s = 10.368 s$

(Inserting the values of ${\alpha}_{F e} = 12 \times {10}^{-} 6 {C}^{-} 1 \mathmr{and} \Delta \theta = {20}^{\circ} C$ )

Sep 12, 2016

We know that for an ideal pendulum time period $T$ is given by the expression
$T = 2 \pi \sqrt{\frac{L}{g}}$
where $L$ is the length of the pendulum and $g$ acceleration due to gravity.
Given that the pendulum keeps correct time at ${20}^{\circ} C$
:.T_"correct"=2pisqrt((L_"20")/g) .......(1)

Now we need to find time period at ${40}^{\circ} C$

We know that with increase in temperature length ${L}_{0}$ of metallic rod expands and the expression is
L_T = L_0 ( 1 + α ΔT) .......(2)
where α# is the coefficient of linear expansion and $\Delta T$ is the change in temperature.

It can be shown that the coefficient of volumetric expansion
$\gamma \approx 3 \alpha$
As such from (2) we get
${L}_{40} = {L}_{20} \left(1 + \frac{36 \times {10}^{-} 6}{3} \times \left(40 - 20\right)\right)$
${L}_{40} = 1.00024 {L}_{20}$ .....(3)

We see that at ${40}^{\circ} C$ length of the pendulum is greater than at ${20}^{\circ} C$.
As the time period is directly proportional to the square root of length, increase in length implies increase in time period. Which amounts to clock loosing time.
From (1)
${T}_{40} = 2 \pi \sqrt{\frac{{L}_{40}}{g}}$
Using (3)
${T}_{40} = 2 \pi \sqrt{\frac{1.00024 {L}_{20}}{g}}$
$\implies {T}_{40} = \sqrt{1.00024} \times {T}_{\text{correct}}$
To calculate loss per day we insert seconds in $1$ day ($24$ hours) as the correct time and deduct seconds equal to one day. We get
$\Delta T i m e = \sqrt{1.00024} \times 86400 - 86400$
$= 10.4 s$, rounded to one decimal place.