# Question #6f534

Oct 6, 2017

Its original length was 0.48 m.

#### Explanation:

The formula that relates period, T, to length, L is
$T = 2 \cdot \pi \cdot \sqrt{\frac{L}{g}}$

Let the original period and length be ${T}_{1} \mathmr{and} {L}_{1}$ respectively. Therefore in the original configuration,
${T}_{1} = 2 \cdot \pi \cdot \sqrt{{L}_{1} / g} \text{ }$ Eq(1)

After the length is increased, the formula looks like this
$1.5 \cdot {T}_{1} = 2 \cdot \pi \cdot \sqrt{\frac{{L}_{1} + 0.6 m}{g}}$

Rearranging the formula
$\sqrt{\frac{{L}_{1} + 0.6 m}{g}} = \frac{1.5 \cdot {T}_{1}}{2 \cdot \pi} = 0.239 \cdot {T}_{1}$
Squaring both sides
$\frac{{L}_{1} + 0.6 m}{g} = 0.057 \cdot {T}_{1}^{2}$

Working to get L_1 by itself
${L}_{1} / g + 0.6 \frac{m}{g} = 0.057 \cdot {T}_{1}^{2}$
${L}_{1} / g = 0.057 \cdot {T}_{1}^{2} - 0.6 \frac{m}{g}$
${L}_{1} = \left(0.057 \cdot {T}_{1}^{2} - 0.6 \frac{m}{g}\right) \cdot g$
${L}_{1} = 0.057 \cdot g \cdot {T}_{1}^{2} - 0.6 m \text{ }$Eq(2)

Now we will substitute the expression for ${T}_{1}$ from Eq(1) in for the ${T}_{1}$ from Eq(2).
${L}_{1} = 0.057 \cdot g \cdot {\left(2 \cdot \pi \cdot \sqrt{{L}_{1} / g}\right)}^{2} - 0.6 m$

Working to simplify that
${L}_{1} = 0.057 \cdot \cancel{g} \cdot {\left(2 \cdot \pi\right)}^{2} \cdot \left({L}_{1} / \cancel{g}\right) - 0.6 m$
${L}_{1} = 0.057 \cdot {\left(2 \cdot \pi\right)}^{2} \cdot \left({L}_{1}\right) - 0.6 m$

${L}_{1} = 2.25 \cdot {L}_{1} - 0.6 m$

Getting close now.
$1.25 {L}_{1} = 0.6 m$

${L}_{1} = \frac{0.6 m}{1.25} = 0.48 m$

I hope this helps,
Steve