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प्रश्न
Compare the time periods of a simple pendulum at places where g = 9.8 ms−2 and 4.36 ms−2 respectively.
संख्यात्मक
उत्तर
g1 = 9.8 ms−2
g2 = 4.36 ms−2
Now, time period of pendulum is, `"T" = 2root(pi)(l/g)`
`"T"_1 = 2root(pi)(l/(g_1))`
`"T"_1 = 2root(pi)(l/9.8)`
= 2π × 0.3194 × 1 = 2.0061
Similarly,
`"T"_2 = 2root(pi)(l/(g_2))`
`"T"_2 = 2root(pi)(l/(4.36))`
= 2π × 0.4789 × 1 = 3.0071
T1 : T2 = 2.0061 : 3.0071
= 0.667 : 1
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