Answer:
[tex]3.331m/s^2[/tex]
Explanation:
Based on our information, the net force on mass [tex]m[/tex] on planet [tex]X[/tex] in the equilibrium position is expressed as:
[tex]F_n_e_t=k\bigtriangleup L-mg_x=0N\\\frac{k}{m}=\frac{g_x}{\bigtriangleup L}[/tex]
#For a simple harmonic motion:
[tex]k/m=(2\pi f)^2[/tex]
Therefore we have:
[tex](2\pi f)^2=\frac{g_x}{\bigtriangleup L}\\\\g_x=(\frac{2\pi}{T})\bigtriangleup L=(\frac{2\pi}{18.2s/11.0})^2 \times 0.231\\\\=3.331m/s^2[/tex]
Hence, the new [tex]g[/tex] is 3.331[tex]m/s^2[/tex]