Question

A simple harmonic oscillator consists of a block of mass$\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\text{kg}}$ attached to a spring of spring constant $\mathbf{100}\mathbf{}\mathbf{\text{N/m}}$. When $\mathbf{\text{t}}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\text{s}}$, the position and velocity of the block are $\mathit{x}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{129}\mathbf{}\mathbf{\text{m}}$ and$\mathit{v}\mathbf{=}\mathbf{}\mathbf{3}\mathbf{.}\mathbf{415}\mathbf{}\mathbf{\text{m}}\mathbf{/}\mathbf{\text{s}}$.

1. What is the amplitude of the oscillations?
2. What was the position of the block at t=0s ?
3. What was the velocity of the block at t=0s ?

Short answer

1. The amplitude of the oscillations is $0.50\text{m}$
2. The position of the block at $\text{t}=0\text{s}$islocalid="1662972713387" $-0.228\text{m}$
3. The velocity of the block at $\text{t}=0\text{s}$is $3.15\text{m}/\text{s}$

Step-by-step solution

Stating the given data

1. Mass of the block,$m=2.00\text{kg}$
2. Spring constant, $k=100\text{N}/\text{m}$
3. At localid="1662972921566" $t=0\text{s},$position $x=0.129\text{m}$, and velocity localid="1662972887523" $v=3.415\text{m}/\text{s}$.

Understanding the concept of motion

Using the relation between angular velocity and spring constant, we can find the angular velocity. Then, using this angular velocity, position function, and velocity function, we can find the amplitude of the oscillations, position, and velocity of the block at $\mathbf{\text{t}}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{\text{s}}$.

Formula:

Angular frequency of a body in oscillation:

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

The general expression for the velocity of motion:

$x=x_m\cos \left(\omega t+\varphi \right)$ ......(ii)

The general expression for the velocity of motion:

$v=-x_m\omega \text{sin}\left(\omega t+\varphi \right)$ ......(iii)

a) Calculation of amplitude

Dividing equation (iii) by (ii), we get

$\begin{array}{rcl}\frac{v}{x}& =& \frac{x_m\omega \sin \left(\omega t+\varphi \right)}{x_m\cos \left(\omega t+\varphi \right)}\\ \frac{v}{x}& =& -\omega \tan \left(\omega t+\varphi \right)\\ \left(\omega t+\varphi \right)& =& {\tan }^{-1}\left(-\frac{v}{\omega x}\right)\\ & & \end{array}$

Using equation (i) and the given values, the angular frequency can be given as:

$\begin{array}{rcl}\omega & =& \sqrt{\frac{100\text{N}/\text{m}}{2.00\text{kg}}}\\ & =& 7.07\text{rad}/\text{s}\end{array}$

Again,

$\begin{array}{rcl}\left(\omega \text{t}+\varphi \right)& =& {\text{tan}}^{-1}\left[-\frac{-3.415\frac{\text{m}}{\text{s}}}{\left(7.07\frac{\text{rad}}{\text{s}}\right)\left(0.129\text{m}\right)}\right]\\ & =& -1.31\text{rad}\end{array}$

Hence, using the above value in equation (ii) and $x=0.129\text{m}$, we get

$\begin{array}{rcl}0.129\text{m}& =& x_m\text{cos}\left(-1.31\right)\\ x_m& =& 0.500\text{m}\end{array}$

Therefore, the amplitude of the oscillations is $0.50\text{m}$

b) Calculation of position at t=0s

At $t=1.00\text{s}$and $\omega =7.07\raisebox{1ex}{$\text{rad}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$using these values in equation (a), we get

$\varphi =-7.38\text{rad}$

From equation (ii), the position can be found as

$\begin{array}{rcl}{x}_0& =& x_m\cos \left(\varphi \right)\\ & =& 0.500\text{m}\text{cos}\left(-7.38\text{rad}\right)\\ & =& -0.228\text{m}\end{array}$

Therefore, the position of the block at $\text{t}=0\text{s}$ is $-0.228\text{m}$

c) Calculation of velocity at t=0s

Using equation (iii) and the values$t=0\text{s},\varphi =-8.38\text{rad},x_m=0.500\text{m}$, we get,

$\begin{array}{rcl}{v}_0& =& \left(-0.500\text{m}\right)\omega \text{sin}\left(-7.38\text{rad}\right)\\ & =& 3.15\text{m}/\text{s}\end{array}$

Therefore, the velocity of the block at $t=0\text{s}$is $3.15\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$.