Question

A block-spring system consists of a spring with constant k = 475 N/m attached to a 2.50 kg block on a frictionless surface. The block is pulled 5.50 cm from equilibrium and released from rest. For the resulting oscillation, find the amplitude, angular frequency, frequency, and period. What is the maximum value of the block's velocity and acceleration?

Answer 1

Explanation:

It is given that,

Spring constant of the spring, k = 475 N/m

Mass of the block, m = 2.5 kg

Elongation in the spring from equilibrium, x = 5.5 cm

(a) We know that the maximum elongation in the spring is called its amplitude. So, the amplitude for the resulting oscillation is 5.5 cm.

(b) Let $\omega$ is the angular frequency. It is given by :

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

$\omega=\sqrt{\dfrac{475}{2.5}}$

$\omega=13.78\ rad/s$

(c) Let T is the period. It is given by :

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

$T=\dfrac{2\pi}{13.78}$

T = 0.45 s

(d) Frequency,

$f=\dfrac{1}{T}$

$f=\dfrac{1}{0.45}$

f = 2.23 Hz

(e) Let v is the maximum value of the block's velocity. It is given by :

$v_{max}=\omega\times A$

$v_{max}=13.78\times 5.5\times 10^{-2}$

$v_{max}=0.757\ m/s$

The value of acceleration is given by :

$a=\omega^2A$

$a=(13.78)^2\times 5.5\times 10^{-2}$

$a=10.44\ m/s^2$

Hence, this is the required solution.