# Question #e1d2a

##### 3 Answers

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

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 vecF_y = vecT - mvecgcostheta = 0 => vecT = mvecgcostheta#

#sum vecF_x = mgsintheta ne 0#

The arc length of travel is

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

So,

#k = (mgsintheta)/(Ltheta)#

For small

#k ~~ (mg)/L#

The period is defined as

#color(blue)(T) = 2pisqrt(m/k)#

#= 2pisqrt(m/((mg)/L)) = 2pisqrt(m*L/(mg)) ~~ color(blue)(2pisqrt(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 = 1/V ((delV)/(delT))_P#

This is saying that **constant-pressure** process in which the **volume** of the substance changes **due to** the change in temperature. *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.**

## 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

**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,

**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,

**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 =2pi sqrt(L/g)#

According to the 2nd law

**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 (

Now by definition linear expansion coefficient

And cubical expansion coefficient

Given

It is also given that pendulum keeps correct time at

Let the effective length of the pendulum be

So

Now by 2nd Law of simple pendulum

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

**In one day** **lose** ) will be given by

(Inserting the values of

We know that for an ideal pendulum time period

where

Given that the pendulum keeps correct time at

Now we need to find time period at

We know that with increase in temperature length

where

It can be shown that the coefficient of volumetric expansion

As such from (2) we get

We see that at

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)

Using (3)

To calculate loss per day we insert seconds in