Question

When a pendulum with a period of 2.00000 s is moved to a new location from one where the acceleration due to gravity was 9.80 m/s2, its new period becomes 1.99782 s. By how much does the acceleration due to gravity differ at its new location?

Answer 1

The new accleration differs in
`0.021\,(m)/(s^(2))` with respect to old location.

Step-by-step explanation:

Let assume that the system is a simple pendulum. The period is:

`T = (2\pi)/(\omega)`

`T = 2\pi \cdot \sqrt {(l)/(g)}`

The following relation is constructed:

`T_(1) \cdot \sqrt{g_(1)} = T_(2) \cdot \sqrt{g_(2)}`

`T_(1)^(2)\cdot g_(1) = T_(2)^(2)\cdot g_(2)`

`g_(2) = \left((T_(1))/(T_(2))\right)^(2)\cdot g_(1)`

The gravity acceleration at new location is:

`g_(2) = \left((2\,s)/(1.99782\,s) \right)^(2)\cdot (9.80\,(m)/(s^(2)) )`

`g_(2) = 9.821\,(m)/(s^(2))`

The new accleration differs in
`0.021\,(m)/(s^(2))` with respect to old location.

Answer 2

0.021
`m/s^2`

Step-by-step explanation:

The period of a pendulum is dependent on the length of the string holding the pendulum, L, and acceleration due to gravity, g. It is given mathematically as:

`T = 2\pi \sqrt{(L)/(g) }`

Let us make L the subject of the formula:

`T^2 = 4\pi ^2((L)/(g)) \\\\\\(L)/(g) = (T^2)/(4\pi ^(2)) \\\\\\L = (gT^2)/(4\pi ^(2))`

We are not told that the length of the string changes, hence, we can conclude that it is constant in both locations.

When the period of the pendulum is 2 s and the acceleration due to gravity is 9.8
`m/s^2`, the length L is:

`L = (9.8 * 2^2)/(4 \pi^(2))\\ \\\\L = 0.9929 m`

When the pendulum is moved to a new location, the period becomes 1.99782 s.

We have concluded that length is constant, hence, we can find the new acceleration due to gravity,
`g_n` :

`0.9929 = (g_n * 1.99782^2)/(4\pi^(2)) \\\\\\0.9929 = 0.1011 g_n`

Therefore:

`g_n = 0.9929/0.1011\\\\\\g_n = 9.821 m/s^2`

The difference between the new acceleration due to gravity,
`g_n` and the former acceleration due to gravity, g, will be:

`g_n - g` =
`9.821 - 9.8` =
`0.021 m/s^2`

The acceleration due to gravity differs by a value of
`0.021 m/s^2` at the new location.