Question

The time period, T, of a simple pendulum is directly proportional to the square root of the length, d, of the pendulum.

When d=6, T=5

Find the value of T when d=3

Input note: give your answer correct to 2 decimal place.

Answer 1

`T = 3.54`

Explanation:

Given

Direct Variation of T to
`\sqrt d`

`d =6;\ when\ T = 5`

Required

Determine T when d = 3

The variation can be represented as:

`T\ \alpha\ \sqrt d`

Convert to equation

`T = k\sqrt d`

`d =6;\ when\ T = 5`; so we have:

`5 = k * \sqrt 6`

Make k the subject:

`k = (5)/(\sqrt 6)`

To solve for T when d = 3.

Substitute 3 for d and
`k = (5)/(\sqrt 6)` in
`T = k\sqrt d`

`T = (5)/(\sqrt 6) * √(3)`

`T = (5√(3))/(\sqrt 6)`

`T = (5 * 1.7321)/(2.4495)`

`T = (8.6605)/(2.4495)`

`T = 3.5356`

`T = 3.54` -- approximated