Question

The free-fall acceleration on Mars is 3.7 m/s2. (a) What length of pendulum has a period of 2.4 s on Earth? cm (b) What length of pendulum would have a 2.4-s period on Mars? cm An object is suspended from a spring with force constant 10 N/m. (c) Find the mass suspended from this spring that would result in a period of 2.4 s on Earth. kg (d) Find the mass suspende

Answer 1

a)
`l=143cm`

b)
`l=54cm`

c)
`m=1.46kg`

d)
`m=1.46kg`

Step-by-step explanation:

The period is defined as:

`T=(2\pi)/(\omega)(1)`

Here,
`\omega` is the natural frequency of the system.

In a pendulum is given by:

`\omega=\sqrt{(g)/(l)}(2)`

Replacing (2) in (1) and solving for l:

`T=2\pi\sqrt{(l)/(g)}\\T^2=4\pi^2(l)/(g)\\l=(gT^2)/(4\pi^2)`

a) On Earth:

`l=(9.8(m)/(s^2)(2.4s)^2)/(4\pi^2)\\l=1.43m=143cm`

b) On Mars:

`l=(3.7(m)/(s^2)(2.4s)^2)/(4\pi^2)\\l=0.54m=54cm`

In a mass-spring system, the natural frequency is:

`\omega=\sqrt{(k)/(m)}(3)`

Replacing (3) in (1) and solving for m:

`T=2\pi\sqrt{(m)/(k)}\\T^2=4\pi^2(m)/(k)\\m=(kT^2)/(4\pi^2)`

c) On Earth:

`m=(10(N)/(m)(2.4s)^2)/(4\pi^2)\\m=1.46kg`

d) On Mars:

`m=(10(N)/(m)(2.4s)^2)/(4\pi^2)\\m=1.46kg`