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
600.3k+ views
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.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

