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

Answer:

0.021 $m/s^2$

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:

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

Let us make L the subject of the formula:

$T^2 = 4\pi ^2(\frac{L}{g}) \\\\\\\frac{L}{g} = \frac{T^2}{4\pi ^{2}} \\\\\\L = \frac{gT^2}{4\pi ^{2}}$

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:

$L = \frac{9.8 * 2^2}{4 \pi^{2}}\\ \\\\L = 0.9929 m$

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, $g_n$ :

$0.9929 = \frac{g_n * 1.99782^2}{4\pi^{2}} \\\\\\0.9929 = 0.1011 g_n$

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.

Answer 2

Answer:

The new accleration differs in $0.021\,\frac{m}{s^{2}}$ with respect to old location.

Explanation:

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

$T = \frac{2\pi}{\omega}$

$T = 2\pi \cdot \sqrt {\frac{l}{g}}$

The following relation is constructed:

$T_{1} \cdot \sqrt{g_{1}} = T_{2} \cdot \sqrt{g_{2}}$

$T_{1}^{2}\cdot g_{1} = T_{2}^{2}\cdot g_{2}$

$g_{2} = \left(\frac{T_{1}}{T_{2}}\right)^{2}\cdot g_{1}$

The gravity acceleration at new location is:

$g_{2} = \left(\frac{2\,s}{1.99782\,s} \right)^{2}\cdot (9.80\,\frac{m}{s^{2}} )$

$g_{2} = 9.821\,\frac{m}{s^{2}}$

The new accleration differs in $0.021\,\frac{m}{s^{2}}$ with respect to old location.