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Thermal Expansion of Solids

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ntThe length of a pendulum is measured as 20.0 cm. The time interval for 100 oscillations is measured as 90s with a stopwatch of 1s resolutions. Find the approximate % change (in g)n

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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. Time period of 10 oscillations of a simple pendulum is found to be 20 sec by a stopwatch of resolution 0.25 sec. Length of the string is measured to be 30 cm by a scale of least count 0.1 cm. Percentage error in measurement of acceleration due to gravity g by simple pendulum is?

T = 2 π \sqrt{\frac{l}{g}}

Q. A student measures the time period of

100

oscillations of a simple pendulum four times. The data set is

90 s, 91 s, 95 s

92 s

. If the minimum division in the measuring clock is

1 s

, then the reported mean time should be:

Q. A student measures the time period of

100

oscillations of a simple pendulum four times. That data set

90 s, 91 s, 95 s

92 s

. If the minimum division in the measuring clock is

1 s,

then he reported time should be

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