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

Answer:

a) $l=143cm$

b) $l=54cm$

c) $m=1.46kg$

d) $m=1.46kg$

Explanation:

The period is defined as:

$T=\frac{2\pi}{\omega}(1)$

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

In a pendulum is given by:

$\omega=\sqrt{\frac{g}{l}}(2)$

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

$T=2\pi\sqrt{\frac{l}{g}}\\T^2=4\pi^2\frac{l}{g}\\l=\frac{gT^2}{4\pi^2}$

a) On Earth:

$l=\frac{9.8\frac{m}{s^2}(2.4s)^2}{4\pi^2}\\l=1.43m=143cm$

b) On Mars:

$l=\frac{3.7\frac{m}{s^2}(2.4s)^2}{4\pi^2}\\l=0.54m=54cm$

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

$\omega=\sqrt{\frac{k}{m}}(3)$

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

$T=2\pi\sqrt{\frac{m}{k}}\\T^2=4\pi^2\frac{m}{k}\\m=\frac{kT^2}{4\pi^2}$

c) On Earth:

$m=\frac{10\frac{N}{m}(2.4s)^2}{4\pi^2}\\m=1.46kg$

d) On Mars:

$m=\frac{10\frac{N}{m}(2.4s)^2}{4\pi^2}\\m=1.46kg$