Question

(a) What is the length of a simple pendulum that oscillates with a period of 1.2 s on Earth, where the acceleration due to gravity is 9.80 m/s2, and on Mars, where the acceleration due to gravity is 3.70 m/s2?
LE = m LM = m
(b) What mass would you need to suspend from a spring with a force constant of 20 N/m in order for the mass-spring system to oscillate with a period of 1.2 s on Earth, where the acceleration due to gravity is 9.80 m/s2, and on Mars, where the acceleration due to gravity is 3.70 m/s2?
mE = kg mM = kg

Answer 1

a) Length of pendulum in Earth = 0.36 m  

   Length of pendulum in Mars = 0.13 m

b) Mass in Earth = Mass in Mars = 0.73 kg

Explanation:

a) Period of simple pendulum

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

   l₁ = length of pendulum in Earth = ?

   l₂ = length of pendulum in Mars = ?

   T = Period of pendulum = 1.2 s

    g₁ = Acceleration due to gravity in Earth = 9.80 m/s²

    g₂ = Acceleration due to gravity in Mars = 3.70 m/s²

For Earth :-

     $T=2\pi \sqrt{\frac{l_1}{g_1}}\\\\1.2=2\pi \sqrt{\frac{l_1}{9.8}}\\\\l_1=0.36m$

     Length of pendulum in Earth = 0.36 m      

For Mars :-

     $T=2\pi \sqrt{\frac{l_2}{g_2}}\\\\1.2=2\pi \sqrt{\frac{l_2}{3.7}}\\\\l_2=0.13m$

     Length of pendulum in Mars = 0.13 m

b) Period of spring

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

 Here period is independent of acceleration due to gravity, so mass value is same in Earth and Mars.

                m = Mass = ?

                 T = Period = 1.2 s

                  k = Spring constant = 20 N/m

              $T=2\pi \sqrt{\frac{m}{k}}\\\\1.2=2\pi \sqrt{\frac{m}{20}}\\\\m=0.73kg$

             Mass in Earth = Mass in Mars = 0.73 kg

Answer 2

The length of a simple pendulum and the mass to suspend the spring are mathematically given as

a L1 = 0.36m for the earth and L2 = 0.13m for mars

b m=0.73kg

What is the length of a simple pendulum?

Question Parameter(s):

period of 1.2 s on Earth

where the acceleration due to gravity is 9.80 m/s2

and on Mars, where the acceleration due to gravity is 3.70 m/s2?

a force constant of 20 N/m

Generally, the equation for the Period of a simple pendulum  is mathematically given as

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

therefore the length of Earth is

$\\\\1.2=2\pi \sqrt{\frac{l_1}{9.8}}\\\\l_1=0.36m$

and for mars

$\\\\1.2=2\pi \sqrt{\frac{l_2}{3.7}}\\\\l_2=0.13m$

In conclusion length of a simple pendulum is

L1 = 0.36m for earth and L2 = 0.13m for mars

What mass would you need to suspend from a spring?

Period of spring

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

$}\\\\1.2=2\pi \sqrt{\frac{m}{20}}\\\\m=0.73kg$

In conclusion, the mass to suspend the spring

m = 0.73kg

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