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Calculate the length of asecond pendulam...

Calculate the length of asecond pendulam at a place where g= 9.8 `ms^(-2)`.
[Hint: A seconds pendulam Is one with a time period of exactly 2 s]

Text Solution

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The correct Answer is:
0.993 m
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Knowledge Check

  • The length of seconds pendulum at a place wl}ere g = 4.9 m/ s^2 is

    A
    99.54 cm
    B
    60 cm
    C
    50 cm
    D
    101 cm

  • The weight of a body is 9.8 N at the place where g=9.8 ms^(-2) . Its mass is

    A
    zero
    B
    9.8 kg
    C
    10 kg
    D
    1 kg

  • The length of a second's pendulum on the surface of the earth, where g=9.8 m//s^(2) , is approximately equal to

    A
    0.8 m
    B
    1 m
    C
    1.2m
    D
    1.1 m

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    What is second's pendulum ? Find the length of second's pendulum for g = 9.8 m//s^(2) ? Also write expressions for the time period if the second's pendulum is an a carriage which is accelerating (a) upwards (b) downwards (c ) horizontally. Also, fin the value of time period of it if is made to oscillate a freely falling lift.

    (i) Calculate the length of a second's pendulum. (ii) If this pendulum is mounted in a lift which accelerates upwards at 2.8 ms^(-2) , by what factor does its period of oscillation change from the original value ? Given g on earth =9.8 ms^(-2) .

    A second's pendulum is taken to a height where the value of g is 4.36ms^(-2) . What will be its new time period?

    A second's pendulum is taken from a place where g=9.8 ms^(-2) to a place where g=9.7 ms^(-2) .How would its length be changed in order that its time period remains unaffected ?

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