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On your first trip to Planet X you happen to take along a 100 g mass, a 40-cm-long spring, a meter stick, and a stopwatch. You're curious about the free-fall acceleration on Planet X, where ordinary tasks seem easier than on earth, but you can't find this information in your Visitor's Guide. One night you suspend the spring from the ceiling in your room and hang the mass from it. You find that the mass stretches the spring by 22.6 cm . You then pull the mass down 10.9 cm and release it. With the stopwatch you find that 10 oscillations take 18.8 s .

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okpalawalter8

Answer:

[tex]g=2.513m/s^2[/tex]

Explanation:

From the question we are told that:

Mass Taken [tex]m=100g \approx 100*10^{-3}[/tex]

Strain on Spring [tex]l_s=22.6cm \approx 22.6*10^{2}[/tex]

Distance pulled down [tex]d_d=10.9[/tex]

Time for ten oscillations take [tex]t=18.8s[/tex]

 

Generally the equation for angular velocity is mathematically given by

[tex]\omega=2\pi\\\omega=2\pi*\frac{10\ oscillations}{t}\\\\\omega=2\pif=2\pi*\frac{10\ oscillations}{18.8}\\Also\\\omega=\sqrt{\frac{k}{m}} \\\omega=\sqrt{\frac{k}{100*10^{-3}}}[/tex]

Therefore

[tex]{2\pi*\frac{10\ oscillations}{18.8}}=\sqrt{\frac{k}{100*10^{-3}}} \\k=\sqrt{2\pi*\frac{10\ oscillations}{18.8}}*100*10^{-3}\\k=1.112N/m[/tex]  

Generally the equation for Acceleration due to gravity is mathematically given by

[tex]F=mg\\mg=k l_s\\g=\frac{k l_s}{m}\\g=\frac{1.112*2.6*10^{2}}{100*10^{-3}}[/tex]

[tex]g=2.513m/s^2[/tex]

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