Question

If a simple pendulum oscillates with small amplitude and its length is doubled, what happens to the frequency of its motion? It doubles. It becomes squareroot 2 times as large. It becomes half as large. It becomes 1/squareroot 2 times as large. It remains the same.

Answer 1

Answer: It becomes
`(1)/(√(2) )` times the original.

Explanation:

Frequency (f)

Length (l)

Gravitational acceleration (g)

f=
`(1)/(2\pi )`
`\sqrt{(g)/(l) }`

Therefore f is indirectly proportional to
`√(l)`.

So, when the length is doubled the frequency becomes
`(1)/(√(2) )` times the original frequency.