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 acceleration due to gravity on Planet X, where ordinary tasks seem easier than on earth, but you can’t find this information in your Visitors 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 23.1 cm . You then pull the mass down 5.40 cm and release it. With the stopwatch you find that 11.0 oscillations take 18.2 s . Can you now satisfy your curiosity?

what is the new g ?


  1. Answer:



    Based on our information, the net force on mass m on planet X in the equilibrium position is expressed as:

    F_n_e_t=k\bigtriangleup L-mg_x=0N\\\frac{k}{m}=\frac{g_x}{\bigtriangleup L}

    #For a simple harmonic motion:

    k/m=(2\pi f)^2

    Therefore we have:

    (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

    Hence, the new g is 3.331m/s^2


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