Problem 417

Sir SagarDecember 28, 2014December 28, 20140

Simple Pendulum:- Gravity is responding for an object falling toward Earth. The farther the object falls, the faster it is moving when it hits the ground. For each second that an object falls, its speed increase by a constant amount, called the acceleration due o gravity, denoted g. One way to calculate the value of g is to use a simple pendulum. See the accompanying figure.

The time T for a pendulum to swing back and forth once is called its period and is given by

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

where L equals the length of the pendulum. The table lists the periods of pendulums with different lengths.

L(feet)	0.5	1.0	1.5
T(seconds)	0.78	1.11	1.36

(a) Solve the formula for g.

(b) Use the table to determine the value of g.(Note: The units for g are feet per second per second.)

(c) Interpret your result.

Solution

A.

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

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

Square both sides

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

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

$T ^{2}\times g = 4\pi ^{2}L$

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

B.

Value of g

when L = 0.5 and T = 0.78

now plugin both values in the part (1) equation

g = $\frac{4\pi^{2}(0.5)}{(0.78)^{2}}$

g = $\frac{2\pi^{2}}{0.6084}$

g = $\frac{2(9.8596)}{0.6084}$

g= $\frac{19.7192}{0.6084}$

g = 32.41 $\frac{ft}{sec^{2}}$

C.

Original value of constant is 9.8 $m/sec^{2}$

We get the value of g from the part (b) is 32.41 $ft/sec^{2}$

1ft = $\frac{1}{3.3}$ meter

So 32.41 $ft/sec^{2}$ = $\frac{32.41}{3.3} m/sec^{2}$

= 9.8 $m/sec^{2}$

Now we can see the both results are same, So we can say we can calculate the value of g by simple pendulum.

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