Question

A 200g block connected to a light spring for which the force constant is 5 N/m is free to oscillate on a frictionless, horizontal surface. The block is displaced 5 cm from equilibrium and released from rest as

1.1. Find the period of motion of the block. (6)

1.2. Determine the maximum speed of the block. (4)

1.3. Determine the maximum acceleration of the block. (4)

1.4. Express the position, velocity, and acceleration as a function of time in SI units.

Answer 1

$m = 200g= 0.2\ kg$

$A = 5cm = 0.05\ m$

$k= 5\ \frac{N}{m}$

$1.1$

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

$T= 2\pi\sqrt{\frac{0.2}{5}}=1.26\ s$

$1.2$

$\text{ maximum } E_p'=\frac{kA^2}{2}$

$\text{ maximum } E_k'=\frac{mv^2_{max}}{2}$

$E'_p=E'_k$

$v_{max}= \sqrt{\frac{kA^2}{m}}$

$v_{max}= \sqrt{\frac{5*0.05^2}{0.2}}=0.25\ \frac{m}{s}$

$1.3$

$F =-kx$

$F=ma$

$a = -\frac{k}{m}x$

$a_{max} \text{ if x =-A}$

$a_{max}=\frac{k}{m}A$

$a_{max}=\frac{5}{0.2}*0.05=1.25\ \frac{m}{s^2}$

$1.4$

$\omega_0= \sqrt{\frac{k}{m}}$

$\omega_0= \sqrt{\frac{5}{0.2}}= 5$

$x(t) = A\cos(\omega_0t)$

$v(t) = x'(t)$

$v(t)=-A \omega_0\sin( \omega_0t)$

$a(t)= v'(t)$

$a(t)=-A\omega^2_0\cos( \omega_0t)$

$x(t) = 0.05\cos(5t)$

$v(t)=-0.25\sin( 5t)$

$a(t)=-1.25\cos( 5t)$

$\text{Answer:}$

$T= 1.26\ s$

$v_{max}= 0.25\ \frac{m}{s}$

$a_{max}=1.25\ \frac{m}{s^2}$

$x(t) = 0.05\cos(5t)$

$v(t)=-0.25\sin( 5t)$

$a(t)=-1.25\cos( 5t)$