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A simple pendulum executing S.H.M. is fa...

A simple pendulum executing S.H.M. is falling freely along with the support. Then

A

Its periodic time decreases

B

Its periodic time increases

C

It does not oscillate at all

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
C
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Knowledge Check

  • The length of simple pendulum executing SHM is increased by 69% The percentage increase in the time period of the pendulum is

    A
    `30%`
    B
    `11%`
    C
    `21%`
    D
    `42%`

  • Simple pendulum is executing simple harmonic motion with time period T . If the length of the pendulum is increased by 21% , then the increase in the time period of the pendulum of the increased length is:

    A
    `22%`
    B
    `13%`
    C
    `50%`
    D
    `10%`

  • The displacement, velocity amplitude of particular executing S.H.M. is related by the expression

    A
    `V= omega sqrt(a^2 - x^2)`
    B
    `V= (a^2 - x^2) omega`
    C
    `v= (a^2 + x^2) omega`
    D
    `V=( sqrt( a^2 + x^2 )) omega`

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    Answer the following questions: (a) Time period of particle is S.H.M. depends on the force constant k and mass m of the particle : T=2pisqrt(m//k). A simple pendulum executes S.H.M. approximately. Why then is the time-period of a pendulum independent of the mass of the pendulum? (b) The motino of simple pendulum is approximatley simple harmonic for small angles of oscillation. For large angle of oscillation, a more involved analysis (beyond the scope of this book) shows that T is greater that 2pisqrt(l//g) . Think of a quanlitative argument to appreciate this result. (c) A man with a wrist watch on his hand falls from the top of tower. Does the watch give correct time during the free fall? (d) What is teh frequency of oscillation of a simple pendulum mounted iin a cabin that is freely falling under gravity?

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