Questions: A 0.141-kg particle undergoes simple harmonic motion along the horizontal x-axis between the points (x1=-0.299 mathrm~m) and (x2=0.387 mathrm~m). The period of oscillation is 0.649 s . Find the frequency, (f), the equilibrium position, (xmathrmeq), the amplitude, (A), the maximum speed, (vmax ), the maximum magnitude of acceleration, (atext max ), the force constant, (k), and the total mechanical energy, (Etext tot ). (f=) (A=) (amax =) (mathrmHz ) (xmathrmeq=) (mathrmm ) (vmax =) (k=) (Etext tot =)

A 0.141-kg particle undergoes simple harmonic motion along the horizontal x-axis between the points (x1=-0.299 mathrm~m) and (x2=0.387 mathrm~m). The period of oscillation is 0.649 s . Find the frequency, (f), the equilibrium position, (xmathrmeq), the amplitude, (A), the maximum speed, (vmax ), the maximum magnitude of acceleration, (atext max ), the force constant, (k), and the total mechanical energy, (Etext tot ). (f=) (A=) (amax =) (mathrmHz ) (xmathrmeq=) (mathrmm ) (vmax =) (k=) (Etext tot =)
Transcript text: A $0.141-\mathrm{kg}$ particle undergoes simple harmonic motion along the horizontal $x$-axis between the points $x_{1}=-0.299 \mathrm{~m}$ and $x_{2}=0.387 \mathrm{~m}$. The period of oscillation is 0.649 s . Find the frequency, $f$, the equilibrium position, $x_{\mathrm{eq}}$, the amplitude, $A$, the maximum speed, $v_{\max }$, the maximum magnitude of acceleration, $a_{\text {max }}$, the force constant, $k$, and the total mechanical energy, $E_{\text {tot }}$. \[ f= \] \[ A= \] \[ a_{\max }= \] \[ \begin{array}{ll} \mathrm{Hz} & x_{\mathrm{eq}}= \\ \mathrm{m} & v_{\max }= \end{array} \] $k=$ $E_{\text {tot }}=$
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Solution

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Solution Steps

Step 1: Calculate the Frequency, \( f \)

The frequency \( f \) is the reciprocal of the period \( T \). Given that the period \( T = 0.649 \) s, the frequency is calculated as:

\[ f = \frac{1}{T} = \frac{1}{0.649} \approx 1.5401 \, \text{Hz} \]

Step 2: Determine the Equilibrium Position, \( x_{\text{eq}} \)

The equilibrium position \( x_{\text{eq}} \) is the midpoint between the maximum and minimum displacements. Given \( x_1 = -0.299 \, \text{m} \) and \( x_2 = 0.387 \, \text{m} \), the equilibrium position is:

\[ x_{\text{eq}} = \frac{x_1 + x_2}{2} = \frac{-0.299 + 0.387}{2} = 0.044 \, \text{m} \]

Step 3: Calculate the Amplitude, \( A \)

The amplitude \( A \) is half the distance between the maximum and minimum displacements:

\[ A = \frac{x_2 - x_1}{2} = \frac{0.387 - (-0.299)}{2} = 0.343 \, \text{m} \]

Final Answer

\[ \boxed{f = 1.5401 \, \text{Hz}} \]

\[ \boxed{x_{\text{eq}} = 0.044 \, \text{m}} \]

\[ \boxed{A = 0.343 \, \text{m}} \]

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