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The length of a pendulum is measured as 20.0cm.The time interval for 100 oscillations is measured as 90 s with a stop watch of 1s resolution.Find the approximate percentage change (in g).

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The period of oscillation of a simple pendulum is $T = 2 π \sqrt{\frac{L}{g}}$ . Measure of $L$ is $20.0 c m$ known to $1 m m$ accuracy and time for $100$ oscillations of the pendulum is found to be $90 s$ using a wrist watch of $1 s$ resolution. The accuracy in the determination of $g$ ?

Q. The period of oscillation of a simple pendulum is

T = 2 π \sqrt{\frac{L}{g}}

. The measured value of

L

20.0 c m

1 m m

accuracy and time for

100

oscillations of the pendulum is found to be

90 s

using a wristwatch of

1 s

resolution. What is the accuracy in the determination of

g

Q. The least count of a stop watch is 0.2 second.The time of 20 oscillations of a pendulum is measured to be 25 seconds.The percentage error in time period is

Q. The least count of a stop-watch is

(1 / 5) s

20

oscillations of a pendulum is measured to be

25 s

. What is the maximum percentage error in this measurement ?

Q. The period of oscillation of a simple pendulum of length l is given by

T = 2 π \sqrt{l / g}

. The length l is about 10 cm and is known to 1 mm accuracy. The period of oscillation is about 0.5 s. The time of 100 oscillations has been measured with a stop watch of 1 s resolution. Find the percentage error in the determination of g.

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