Question

A simple pendulum performs simple harmonic motion about x = 0 with an amplitude A and time period T. The speed of the pendulum at x =A/2 will be ____________.

Options

• `pi "A"  sqrt3 //"T"`

• `pi  "A"//"T"`

• `sqrt3/2 pi"A"//"T"`

• `3pi^2  "A"//"T"`

Answer 1

A simple pendulum performs simple harmonic motion about x = 0 with an amplitude A and time period T. The speed of the pendulum at x =A/2 will be `pi "A"  sqrt3 //"T"`.

Explanation:

`"v" = omega sqrt("A"^2 - "x"^2), "x" = "A"/2`

`therefore "v" = omega sqrt ("A"^2 - "A"^2/4)`

`therefore omega sqrt(3/4 "A") = (omega "A")/2 sqrt3`

`because omega = (2pi)/"T"`

`therefore "v" = (2pi)/(2"T") "A" sqrt3`

` = (pi"A" sqrt3)/"T"`