
In a pendulum, time period is measured by $0.2\% $ accuracy and length is measured by $0.5\% $ accuracy. Find the percentage accuracy in the value of g.
(A) $0.3\% $
(B) $0.9\% $
(C) $0.7\% $
(D) $0.1\% $
Answer
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Hint
First establish the equation of motion of a simple pendulum. From that expression derive the value of the time-period. From that substitute the given values in the suitable formula to calculate the percentage accuracy.
$\Rightarrow T = 2\Pi {\left( {\dfrac{l}{g}} \right)^{1/2}}$ where $T$ is the time period of oscillation, $l$ is the length of the pendulum and $g$ is the acceleration due to gravity.
$\Rightarrow \dfrac{{\Delta g}}{g} = \dfrac{{\Delta L}}{L} + \dfrac{{2\Delta T}}{T}$
Complete step by step answer
First let us learn what a simple pendulum is and how it works. A simple pendulum consists of a heavy point mass suspended from a rigid support by means of an inextensible, weightless and perfectly flexible string.
The equation of motion of the pendulum can be written as,
$\Rightarrow ml\dfrac{{{\partial ^2}\theta }}{{\partial {t^2}}} = - mg\sin \theta $
When $\theta $ is small, we have
$\Rightarrow \dfrac{{{\partial ^2}\theta }}{{\partial {t^2}}} + {\omega ^2}\theta = 0$ where $\Rightarrow \omega = {\left( {\dfrac{g}{l}} \right)^{1/2}}$
Which is the characteristic differential equation for simple harmonic motion.
Therefore, $\dfrac{{2\Pi }}{T} = {\left( {\dfrac{g}{l}} \right)^{1/2}}$
$\Rightarrow T = 2\Pi {\left( {\dfrac{g}{l}} \right)^{1/2}}$ where $T$ is the time-period of oscillation of the pendulum.
Now, given $\dfrac{{\Delta T}}{T} = 0.2$ and $\dfrac{{\Delta l}}{l} = 0.5$
Since $ \Rightarrow T = 2\Pi {\left( {\dfrac{g}{l}} \right)^{1/2}}$
$\Rightarrow \dfrac{{\Delta g}}{g} = \dfrac{{\Delta L}}{L} + \dfrac{{2\Delta T}}{T}$
$\Rightarrow \dfrac{{\Delta g}}{g} = 0.2 + 0.5$
$\Rightarrow \dfrac{{\Delta g}}{g} = 0.9$
Therefore, the percentage accuracy in the value of g is $0.9\% $
So, the correct option is (B).
Additional Information
The motion of a simple pendulum is based upon the following laws:-
- The law of isochronism-This means that the pendulum takes equal time to complete each oscillation.
- The law of length- $T \propto \sqrt l $, when $g$ is constant.
- The law of acceleration due to gravity- $T \propto \dfrac{1}{{\sqrt g }}$, when $l$ is constant
- The law of mass- The time period of oscillation of a pendulum is independent of the mass and material of the bob.
Note
All the laws stated above are only applicable in ideal cases. In reality the pendulum undergoes damped vibration which opposes its motion and thus after a certain amount of time it comes to rest. For the pendulum to undergo periodic motion in such a case, external force has to be applied.
First establish the equation of motion of a simple pendulum. From that expression derive the value of the time-period. From that substitute the given values in the suitable formula to calculate the percentage accuracy.
$\Rightarrow T = 2\Pi {\left( {\dfrac{l}{g}} \right)^{1/2}}$ where $T$ is the time period of oscillation, $l$ is the length of the pendulum and $g$ is the acceleration due to gravity.
$\Rightarrow \dfrac{{\Delta g}}{g} = \dfrac{{\Delta L}}{L} + \dfrac{{2\Delta T}}{T}$
Complete step by step answer
First let us learn what a simple pendulum is and how it works. A simple pendulum consists of a heavy point mass suspended from a rigid support by means of an inextensible, weightless and perfectly flexible string.
The equation of motion of the pendulum can be written as,
$\Rightarrow ml\dfrac{{{\partial ^2}\theta }}{{\partial {t^2}}} = - mg\sin \theta $
When $\theta $ is small, we have
$\Rightarrow \dfrac{{{\partial ^2}\theta }}{{\partial {t^2}}} + {\omega ^2}\theta = 0$ where $\Rightarrow \omega = {\left( {\dfrac{g}{l}} \right)^{1/2}}$
Which is the characteristic differential equation for simple harmonic motion.
Therefore, $\dfrac{{2\Pi }}{T} = {\left( {\dfrac{g}{l}} \right)^{1/2}}$
$\Rightarrow T = 2\Pi {\left( {\dfrac{g}{l}} \right)^{1/2}}$ where $T$ is the time-period of oscillation of the pendulum.
Now, given $\dfrac{{\Delta T}}{T} = 0.2$ and $\dfrac{{\Delta l}}{l} = 0.5$
Since $ \Rightarrow T = 2\Pi {\left( {\dfrac{g}{l}} \right)^{1/2}}$
$\Rightarrow \dfrac{{\Delta g}}{g} = \dfrac{{\Delta L}}{L} + \dfrac{{2\Delta T}}{T}$
$\Rightarrow \dfrac{{\Delta g}}{g} = 0.2 + 0.5$
$\Rightarrow \dfrac{{\Delta g}}{g} = 0.9$
Therefore, the percentage accuracy in the value of g is $0.9\% $
So, the correct option is (B).
Additional Information
The motion of a simple pendulum is based upon the following laws:-
- The law of isochronism-This means that the pendulum takes equal time to complete each oscillation.
- The law of length- $T \propto \sqrt l $, when $g$ is constant.
- The law of acceleration due to gravity- $T \propto \dfrac{1}{{\sqrt g }}$, when $l$ is constant
- The law of mass- The time period of oscillation of a pendulum is independent of the mass and material of the bob.
Note
All the laws stated above are only applicable in ideal cases. In reality the pendulum undergoes damped vibration which opposes its motion and thus after a certain amount of time it comes to rest. For the pendulum to undergo periodic motion in such a case, external force has to be applied.
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