Question

Block with m = 300 grams oscillates at the end of a linear spring with k = 6.5 N/m. (assume this is a top-down view and the block is sliding on a frinctionless surface; neglect gravity)
a) what is the period of the block's motion in seconds

b) if the blovk's max acceleration is 2 m/s^2, what is the amplitude of the motion (m)

Answer 1

Explanation:

Given that,

Mass of the block, m = 300 g = 0.3 kg

Linear spring constant, k = 6.5 N/m

(a) Let T is the period of the block's motion. It is given by :

$T=\dfrac{2\pi}{\omega}$

where

$\omega=\sqrt{\dfrac{k}{m}}$= angular frequency

Also, $\omega=\sqrt{\dfrac{k}{m}}$

$T=\dfrac{2\pi}{\sqrt{\dfrac{k}{m}}}$

$T=\dfrac{2\pi}{\sqrt{\dfrac{6.5}{0.3}}}$

T = 1.34 seconds

(b) The maximum acceleration of the block is, $a_{max}=2\ m/s^2$

The maximum acceleration is given by :

$a_{max}=\omega^2A$

A is the amplitude of the motion,

$A=\dfrac{a_{max}}{\omega^2}$

$A=\dfrac{a_{max}}{(\sqrt{\dfrac{k}{m}})^2}$

$A=\dfrac{ma_{max}}{k}$

$A=\dfrac{0.3\times 2}{6.5}$

A = 0.09 meters

Hence, this is the required solution.