Question

A block oscillating on a spring has a periodT= 2s. What is the period if:

(a) The block’s mass is doubled? Explain. Note that you do not know that value of eithermork, so donotassume any particular values for them. The required analysis involvesthinkingabout ratios.

(b) The value of the spring constant is quadrupled?

(c) The oscillation amplitude is doubled whilemandkare unchanged

Answer 1

(a)

T=2
`\pi`
`√(m/k)`

Step-by-step explanation:

T=2
`\pi`
`√(2m/k)`

=2
`\pi`
`√(m/k) √(2)`

=2
`√(2)`

=2.83 seconds

(b)

T=2 x
`1/√(4)`

=2/2

T=1 second

(c)

T= 2 seconds

T does not depend on the amplitude

Answer 2

Step-by-step explanation:

The period of a mass-spring system is defined as:

`T=2\pi\sqrt(m)/(k)`

Here m is the block's mass and k is the spring constant

(a) We have
`m'=2m`. So:

`T'=2\pi\sqrt(m')/(k)\\T'=2\pi\sqrt(2m)/(k)\\T'=(√(2))2\pi\sqrt(m)/(k)\\T'=(√(2))T`

(b) We have
`k'=4k`. So:

`T'=2\pi\sqrt(m)/(k')\\T'=2\pi\sqrt(m)/(4k)\\T'=((1)/(2))2\pi\sqrt(m)/(k)\\T'=((1)/(2)})T`

(c) The period does not depend on the oscillation amplitude, so we have the same period in both cases.