
A student uses a simple pendulum of exactly 1 m length to determine $g$, the acceleration due to gravity. He uses a stopwatch with the least count of 1 sec for this and records 40 seconds for 20 oscillations. For this observation, which of the following statements is true?
A. Error $\Delta T$ in measuring $T$, the time period, is 0.05 seconds
B. Error $\Delta T$ in measuring $T$, the time period, is 1 second
C. Percentage error in the determination of $g$ is 5%
D. Both A and C
Answer
554.4k+ views
Hint: Find the time period and use it to determine the error in it. Find the percentage error in $g$ using the formula for the time period of a simple pendulum.
Formula used: $T=2\pi \sqrt{\dfrac{l}{g}}$
$T$ is the time period of the pendulum
$l$ is the length of the pendulum
$g$ is the acceleration due to gravity.
Complete step by step answer:
20 oscillations take 40 seconds
1 oscillation takes 2 seconds
Therefore, the time period
$T=2$ seconds
Now,
\[\dfrac{\Delta T}{T}=\dfrac{\Delta t}{t}=\dfrac{1}{40}\]
\[\Delta T=0.05\]seconds
Squaring and rearranging the formula for time period,
$g=\dfrac{4{{\pi }^{2}}l}{{{T}^{2}}}$
Taking log on both sides,
$\log g=\log 4{{\pi }^{2}}l-2\log T$
Differentiating both sides ($4{{\pi }^{2}}l$ is constant and error is additive),
$
\dfrac{\Delta g}{g}=\dfrac{2\Delta T}{T} \\
\dfrac{\Delta g}{g}=\dfrac{2\times 0.05}{2}=0.05 \\
$
The percentage error in $g$ is 5%
Hence, both options A and C are correct
So, the correct answer is “Option D”.
Additional Information:
If two simple pendulums are connected by a spring, then this coupled pendula will have two fundamental frequencies corresponding to compression and elongation of the spring.
A spherical pendulum moves in 3 dimensions on the surface of a hypothetical sphere centred about the pivot point.
Note:
The simple pendulum executes simple harmonic motion due to restoring gravitational force which tends to accelerate it towards its equilibrium position. The simple pendulum is a simplified model of a real physical pendulum by making certain assumptions. The string is massless and inextensible and the bob is a point mass. Motion is restricted in 2 dimensions. Friction and other processes causing energy loss are neglected.
Formula used: $T=2\pi \sqrt{\dfrac{l}{g}}$
$T$ is the time period of the pendulum
$l$ is the length of the pendulum
$g$ is the acceleration due to gravity.
Complete step by step answer:
20 oscillations take 40 seconds
1 oscillation takes 2 seconds
Therefore, the time period
$T=2$ seconds
Now,
\[\dfrac{\Delta T}{T}=\dfrac{\Delta t}{t}=\dfrac{1}{40}\]
\[\Delta T=0.05\]seconds
Squaring and rearranging the formula for time period,
$g=\dfrac{4{{\pi }^{2}}l}{{{T}^{2}}}$
Taking log on both sides,
$\log g=\log 4{{\pi }^{2}}l-2\log T$
Differentiating both sides ($4{{\pi }^{2}}l$ is constant and error is additive),
$
\dfrac{\Delta g}{g}=\dfrac{2\Delta T}{T} \\
\dfrac{\Delta g}{g}=\dfrac{2\times 0.05}{2}=0.05 \\
$
The percentage error in $g$ is 5%
Hence, both options A and C are correct
So, the correct answer is “Option D”.
Additional Information:
If two simple pendulums are connected by a spring, then this coupled pendula will have two fundamental frequencies corresponding to compression and elongation of the spring.
A spherical pendulum moves in 3 dimensions on the surface of a hypothetical sphere centred about the pivot point.
Note:
The simple pendulum executes simple harmonic motion due to restoring gravitational force which tends to accelerate it towards its equilibrium position. The simple pendulum is a simplified model of a real physical pendulum by making certain assumptions. The string is massless and inextensible and the bob is a point mass. Motion is restricted in 2 dimensions. Friction and other processes causing energy loss are neglected.
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