Question

The free-fall acceleration on Mars is 3.7 m/s² 

(a) What length pendulum has a period of 2.0-s on Earth? 

(b) What length pendulum would have a 2.0-s period on Mars? 

An object is suspended from a spring with force constant 10 N/m. 

(c) Find the mass suspended from this spring that would result in a 2.0 s period on Earth. 

(d) Find the mass suspended from this spring that would result in a 2.0 s period on Mars.

Answer 1

Concepts and reason

The concepts used in this problem isthe period of the simple harmonic motion.

First, obtain the expression for the length of the pendulum from the expression of its period. Then calculate the value of length using given values of acceleration of the earth and the Mars.

Finally, obtain the expression for the mass of the spring-mass system from the expression of a period of the spring-mass system. Laterusegiven values of its period and spring constant in the same expression to calculate the value of the mass.

Fundamentals

A motion that repeats itself in equal intervals of time is periodic or simple harmonic motion.

Simple pendulum and spring-mass system are examples of simple harmonic motion.

The period of the simple pendulum is expressed as follows,

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

Here, period the pendulum is $T$ , length of the pendulum is $L$ and acceleration due to gravity is $g$ .

The period of the simple pendulum is expressed as follows,

$T = 2\pi \sqrt {\frac{m}{k}}$

Here, period the spring-mass system is $T$ , the mass of the spring-mass system is $m$ and spring constant is $k$ .

(a.1)

Period of the simple pendulum is,

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

Rewrite the above expression for the length, period and gravitational acceleration on the earth.

${T_{Earth}} = 2\pi \sqrt {\frac{{{L_{Earth}}}}{{{g_{Earth}}}}}$

Rearrange the above equation only for ${L_{Earth}}$ .

${L_{Earth}} = \frac{{{g_{Earth}}{T_{Earth}}^2}}{{4{\pi ^2}}}$

Substitute, $2.0\,{\rm{s}}$ for ${T_{Earth}}$ , $9.8\,{\rm{m/}}{{\rm{s}}^{\rm{2}}}$ for ${g_{Earth}}$ in the above equation.

$\begin{array}{l}\\{L_{Earth}} = \frac{{\left( {9.8\frac{{\rm{m}}}{{{{\rm{s}}^{\rm{2}}}}}} \right){{\left( {2.0\,{\rm{s}}} \right)}^2}}}{{4{{\left( {3.14} \right)}^2}}}\\\\{L_{Earth}} = 0.99\,{\rm{m}}\\\end{array}$

(a.2)

Period of the simple pendulum is,

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

Rewrite the above expression for the length, period and gravitational acceleration on the Mars.

${T_{Mars}} = 2\pi \sqrt {\frac{{{L_{Mars}}}}{{{g_{Mars}}}}}$
‎

Rearrange the above equation only for ${L_{Mars}}$ .

${L_{Mars}} = \frac{{{g_{Mars}}{T_{Mars}}^2}}{{4{\pi ^2}}}$

Substitute, $2.0\,{\rm{s}}$ for ${T_{Mars}}$ , $3.7\,{\rm{m/}}{{\rm{s}}^{\rm{2}}}$ for ${g_{Mars}}$ in the above equation.

$\begin{array}{l}\\{L_{Mars}} = \frac{{\left( {3.7\frac{{\rm{m}}}{{{{\rm{s}}^{\rm{2}}}}}} \right){{\left( {2.0\,{\rm{s}}} \right)}^2}}}{{4{{\left( {3.14} \right)}^2}}}\\\\{L_{Mars}} = 0.37\,{\rm{m}}\\\end{array}$

(b.1)

Period of the spring-mass system is,

$T = 2\pi \sqrt {\frac{m}{k}}$

Rewrite the above expression for mass, spring constant and period for Earth. Spring constant is a constant hence it need not to be metioned for Earth separately.

${T_{Earth}} = 2\pi \sqrt {\frac{{{m_{Earth}}}}{k}}$

Rearrange the above equation only for ${m_{Earth}}$ .

${m_{Earth}} = \frac{{k{T_{Earth}}^2}}{{4{\pi ^2}}}$

Substitute, $2.0\,{\rm{s}}$ for ${T_{Earth}}$ , $10\,{\rm{N/m}}$ for $k$ and $3.14$ for $\pi$ in the above equation.

$\begin{array}{l}\\{m_{Earth}} = \frac{{\left( {10\,{\rm{N/m}}} \right){{\left( {2.0\,{\rm{s}}} \right)}^2}}}{{4{{\left( {3.14} \right)}^2}}}\\\\{m_{Earth}} = 1.01\,{\rm{kg}}\\\end{array}$

(b.2)

Period of the spring-mass system is,

$T = 2\pi \sqrt {\frac{m}{k}}$

Rewrite the above expression for mass, spring constant and period for Mars. Spring constant is a constant hence it need not to be metioned for Mars separately.

${T_{Mars}} = 2\pi \sqrt {\frac{{{m_{Mars}}}}{k}}$

Rearrange the above equation only for ${m_{Mars}}$ .

${m_{Mars}} = \frac{{k{T_{Mars}}^2}}{{4{\pi ^2}}}$

Substitute, $2.0\,{\rm{s}}$ for ${T_{Mars}}$ , $10\,{\rm{N/m}}$ for $k$ and $3.14$ for $\pi$ in the above equation.

$\begin{array}{l}\\{m_{Mars}} = \frac{{\left( {10\,{\rm{N/m}}} \right){{\left( {2.0\,{\rm{s}}} \right)}^2}}}{{4{{\left( {3.14} \right)}^2}}}\\\\{m_{Mars}} = 1.01\,{\rm{kg}}\\\end{array}$

Ans: Part a.1

Length of the pendulum on the earth is ${\bf{0}}{\bf{.99}}\,{\bf{m}}$ .

Part a.2

Length of the pendulum on Mars is ${\bf{0}}.37\,{\bf{m}}$ .

Part b.1

The mass suspended from the spring on Earth is ${\bf{1}}{\bf{.01}}\,{\bf{kg}}$ .

Part b.2

The mass suspended from the spring on Mars is ${\bf{1}}{\bf{.01}}\,{\bf{kg}}$ .