
A simple pendulum has a period $T$ on the earth what is the period ${{T}^{1}}$ of this same pendulum on the moon, where the acceleration due to gravity is $\dfrac{1}{6}$ that of earth.
A)$\dfrac{T}{\sqrt{6}}$
B)$\dfrac{T}{6}$
C)$\sqrt{6T}$
D)$36T$
Answer
492.9k+ views
Hint : Acceleration depends on the mass and also on the force. The force velocity, acceleration and momentum have both magnitude and a direction. Heavier objects have less acceleration compared to lighter objects. Gravitation is a natural phenomenon in which objects with mass or energy are attracted towards other objects. Due to moon gravity tides are formed in the ocean and on earth gravity will give mass to objects.
Complete step-by-step solution:
Simple pendulum is considered to be a point mass suspended from a string or rod of negligible mass.
The time period of a pendulum depends on the square root of the length of the string of the pendulum from its point of suspension and the square root of the effective downward vertical acceleration.
Time period of a simple pendulum (T) is given by:
$T=2\pi \sqrt{\dfrac{L}{g}}$
Where $L$ = pendulum Length.
$g$ = acceleration due to gravity.
On Earth the time period of simple pendulum is:
$T=2\pi \sqrt{\dfrac{L}{{{g}_{e}}}}$ $\cdots \cdots (1)$
Where ${{g}_{e}}$ = acceleration due to gravity of the earth.
On moon the time period of simple pendulum is :
${{T}^{1}}=2\pi \sqrt{\dfrac{L}{{{g}_{m}}}}$ $\cdots \cdots (2)$
${{g}_{m}}$ = acceleration due to gravity of the moon
From the data ${{g}_{m}}=\dfrac{1}{6}{{g}_{e}}$
After substitution
${{T}^{1}}=2\pi \sqrt{\dfrac{L}{\dfrac{1}{6}{{g}_{e}}}}$
Dividing equation (2) by equation (1) then we get
${{T}^{1}}=\sqrt{6T}$
So the correct option is C.
Note:The effect of acceleration on the time period of a pendulum is such that an upward acceleration makes the time period smaller while acceleration in the downward direction makes the time period larger. Changing gravity also has an effect on the time period and a larger gravitational force makes the time period shorter. Thus a pendulum will have a larger time period on the moon where the gravity is less than on earth.
Complete step-by-step solution:
Simple pendulum is considered to be a point mass suspended from a string or rod of negligible mass.
The time period of a pendulum depends on the square root of the length of the string of the pendulum from its point of suspension and the square root of the effective downward vertical acceleration.
Time period of a simple pendulum (T) is given by:
$T=2\pi \sqrt{\dfrac{L}{g}}$
Where $L$ = pendulum Length.
$g$ = acceleration due to gravity.
On Earth the time period of simple pendulum is:
$T=2\pi \sqrt{\dfrac{L}{{{g}_{e}}}}$ $\cdots \cdots (1)$
Where ${{g}_{e}}$ = acceleration due to gravity of the earth.
On moon the time period of simple pendulum is :
${{T}^{1}}=2\pi \sqrt{\dfrac{L}{{{g}_{m}}}}$ $\cdots \cdots (2)$
${{g}_{m}}$ = acceleration due to gravity of the moon
From the data ${{g}_{m}}=\dfrac{1}{6}{{g}_{e}}$
After substitution
${{T}^{1}}=2\pi \sqrt{\dfrac{L}{\dfrac{1}{6}{{g}_{e}}}}$
Dividing equation (2) by equation (1) then we get
${{T}^{1}}=\sqrt{6T}$
So the correct option is C.
Note:The effect of acceleration on the time period of a pendulum is such that an upward acceleration makes the time period smaller while acceleration in the downward direction makes the time period larger. Changing gravity also has an effect on the time period and a larger gravitational force makes the time period shorter. Thus a pendulum will have a larger time period on the moon where the gravity is less than on earth.
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