Q5 E - University Physics with Modern Physics

Question

A machine part is undergoing SHM with a frequency of $\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{ }\mathit{H}\mathit{z}$ and amplitude $\mathbf{1}\mathbf{.}\mathbf{80}\mathbf{ }\mathit{c}\mathit{m}$. How long does it take the part to go from $\mathit{x}\mathbf{=}\mathbf{0}$ to $\mathit{x}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{80}\mathbf{ }\mathit{c}\mathit{m}$?

Step-by-Step Solution

Answer

The machine part takes $62.5 ms$ to go from $x=0$  to $x=-1.80 cm$.

 

Step 1: Definition of amplitude and frequency

The amplitude of motion can be defined as the maximum displacement that an object travels from the equilibrium position.

 

The frequency is defined as the number of oscillations done by an object per second.

Step 2: Calculate the time period.

Consider the given data as below.

The frequency,  $f=4.00Hz$  

 

The relation between time period and frequency is given by,

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

Here, $T$ is the time period and $f$ is the frequency.

 

Substitute the value of frequency to get the value of time period

$$\begin{gathered} T=\frac{1}{4.00Hz} \\ =0.25 s \end{gathered}$$

Step 3: Calculate the time taken by the part to travel from x = 0 to x = - 1 . 80   c m

Consider the amplitude of motion is $A$. So, from $+A$ to $-A$ and back to $+A$ , or from  $0$ to +A , back through 0 to $-A$ and back to  $0$ form a complete vibration or cycle as shown in figure below.