Answer:
[tex]T_{mars}=3.04 s[/tex]
Explanation:
given,
Time period of pendulum on earth = 1.85 s
acceleration of gravity of mars about that of earth = 0.37
Time period of mars = ?
using equation of time period
[tex]T =2\pi\sqrt{\dfrac{l}{g}}[/tex]
now,
time period of mars
[tex]T_{mars} =2\pi\sqrt{\dfrac{l}{g_{mars}}}[/tex]..........(1)
time period of earth
[tex]T_{earth} =2\pi\sqrt{\dfrac{l}{g_{earth}}}[/tex]..........(2)
dividing equation (2)/ (1)
[tex]\dfrac{T_{earth}}{T_{mars}}=\sqrt{\dfrac{g_{mars}}{g_{earth}}}[/tex]
[tex]\dfrac{T_{earth}}{T_{mars}}=\sqrt{\dfrac{0.37g_{earth}}{g_{earth}}}[/tex]
[tex]\dfrac{1.85}{T_{mars}}=\sqrt{0.37}[/tex]
[tex]T_{mars}=\dfrac{1.85}{\sqrt{0.37}}[/tex]
[tex]T_{mars}=3.04 s[/tex]
Time period of pendulum on mars is equal to 3.04 s
The period of the pendulum on mars is 3.03 s.
The given parameters;
The period of the pendulum on mars is calculated as follows;
[tex]T = 2\pi \sqrt{\frac{l}{g} } \\\\T = \frac{k}{\sqrt{g} } \\\\T_1\sqrt{g_1} = T_2\sqrt{g_2} \\\\T_e\sqrt{g_e} = T_m\sqrt{g_m} \\\\T_m = \frac{T_e\sqrt{g_e}}{\sqrt{g_m} } \\\\T_m = T_e \sqrt{\frac{g_e}{g_m} } \\\\T_m = 1.85 \times \sqrt{\frac{g}{0.37g} } \\\\T_m = 3.03 \ s[/tex]
Thus, the period of the pendulum on mars is 3.03 s.
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