Question

An object of mass \( 0.2 \, \text{kg} \) executes simple harmonic motion along the \( x \)-axis with a frequency of \( \left( \frac{25}{\pi} \right) \, \text{Hz} \). At the position \( x = 0.04 \, \text{m} \), the object has kinetic energy \( 0.5 \, \text{J} \) and potential energy \( 0.4 \, \text{J} \). The amplitude of oscillation is \(\dots\) \(\text{cm}\).

Answer 1

Correct Answer: 6

In simple harmonic motion (SHM), total mechanical energy \(E\) equals the sum of kinetic energy (KE) and potential energy (PE). With \(KE = 0.5 \, \text{J}\) and \(PE = 0.4 \, \text{J}\), the total energy is:

\[E = KE + PE = 0.5 + 0.4 = 0.9 \, \text{J}\]

The formula for total mechanical energy in SHM is also expressed as:

\[E = \frac{1}{2} m \omega^2 A^2\]

Here, \(m = 0.2 \, \text{kg}\) represents mass, \(\omega\) is angular frequency, and \(A\) is amplitude. Given a frequency \(f\) of \(\frac{25}{\pi} \, \text{Hz}\), the angular frequency \(\omega\) is calculated as:

\[\omega = 2\pi f = 2\pi \left(\frac{25}{\pi}\right) = 50 \, \text{rad/s}\]

Substituting these values into the energy equation yields:

\[0.9 = \frac{1}{2} \cdot 0.2 \cdot 50^2 \cdot A^2\]

Solving for \(A^2\):

\[0.9 = 0.1 \cdot 2500 \cdot A^2\]

\[0.9 = 250 \cdot A^2\]

Dividing both sides by 250 gives:

\[A^2 = \frac{0.9}{250}\]

\[A^2 = 0.0036\]

Taking the square root to determine \(A\):

\[A = \sqrt{0.0036} = 0.06 \, \text{m}\]

Converting to centimeters:

\[A = 0.06 \times 100 = 6 \, \text{cm}\]

The calculated amplitude is \(6\) cm, consistent with the range provided: 6,6.