# A pendulum has a period of 0.80 s on Earth. What is its period on Mars, where the acceleration of gravity is about 0.37 that on Earth?

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Jun 13, 2016

${T}_{m} \cong 0.75 \text{ } s$

#### Explanation:

${T}_{e} = 2 \pi \sqrt{\frac{l}{g} _ e} \text{ period equation for earth}$

${T}_{m} = 2 \pi \sqrt{\frac{l}{g} _ m} \text{ period equation for mars}$

$\frac{{T}_{e}}{{T}_{m}} = \frac{\cancel{2 \pi} \sqrt{\frac{l}{g} _ e}}{\cancel{2 \pi} \sqrt{\frac{l}{g} _ m}}$

$\text{ plug } {T}_{e} = 0.80 s$

(0.80)/(T_m)=sqrt((l/(g_e))/(l/(g_m)))"

 (0.80)/(T_m)=sqrt(cancel(l)/(g_e)*(g_m)/cancel(l)

$\frac{0.80}{T} _ m = \sqrt{\frac{{g}_{m}}{{g}_{e}}}$

$\text{ plug " g_e=g" } {g}_{m} = 0.37 g$

$\frac{0.80}{{T}_{m}} = \sqrt{\frac{0.37 \cdot \cancel{g}}{\cancel{g}}}$

$\frac{0.80}{T} _ m = \sqrt{0.37}$

${\left(\frac{0.80}{T} _ m\right)}^{2} = {\left(\sqrt{0.37}\right)}^{2}$

$\frac{0.64}{T} _ {m}^{2} = 0.37$

${T}_{m}^{2} = \frac{0.37}{0.64}$

${T}_{m} = \sqrt{\frac{0.37}{0.64}}$

${T}_{m} \cong \frac{0.6}{0.8}$

${T}_{m} \cong \frac{6}{8} = \frac{3}{4}$

${T}_{m} \cong 0.75 \text{ } s$

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