Answer:
c. T
and the new period is [tex]T'=4T[/tex]
Explanation:
the period of a pendulum is given by:
[tex]T=2\pi \sqrt{\frac{l}{g} }[/tex]
as we can see the period of oscillation of the pendulum does not depend on the mass, so if the mass is reduced this will not affect the period, it will remain the same.
The new period if thr pedulum oscilates on Pluto, will be calculated with the previous equation, but the gravity will be
[tex]g'=\frac{1}{16}g[/tex]
where g is the gravity on earth: [tex]g=9.8m/s^2[/tex]
so the new period:
[tex]T=2\pi\sqrt{\frac{l}{g'} } =2\pi\sqrt{\frac{l}{\frac{1}{16} g} } =2\pi\sqrt{\frac{16l}{ g} } =2\pi \sqrt{16}\sqrt{\frac{l}{ g} } =4(2\pi \sqrt{\frac{l}{ g} })[/tex]
and we know that
[tex](2\pi \sqrt{\frac{l}{ g} })[/tex] is the original period [tex]T[/tex]
so the new period in terms of the orginal period is:
[tex]T'=4T[/tex]
The new period oscillation will be "4T".
According to the question,
We know that,
In Earth,
→ [tex]T = 2r \sqrt{\frac{L}{9} }[/tex]
In Pluto,
→ [tex]g_p = \frac{g}{16}[/tex]
∴ [tex]T_p = 2r \sqrt{\frac{L}{\frac{9}{16} } }[/tex]
[tex]= 8r \sqrt{\frac{L}{9}}[/tex]
[tex]= 4T[/tex]
Thus the answer above is right.
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