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

The total mechanical energy \(E\) in simple harmonic motion (SHM) is the sum of kinetic energy (KE) and potential energy (PE). Given \(KE = 0.5 \, \text{J}\) and \(PE = 0.4 \, \text{J}\), we have:

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

The total mechanical energy in SHM is also given by:

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

Where \(m = 0.2 \, \text{kg}\) is the mass, \(\omega\) is the angular frequency, and \(A\) is the amplitude. The frequency \(f\) is given as \(\frac{25}{\pi} \, \text{Hz}\). Therefore, the angular frequency \(\omega\) is:

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

Substitute the values into the energy equation:

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

Solve for \(A^2\):

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

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

Divide both sides by 250:

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

\[A^2 = 0.0036\]

Take the square root to find \(A\):

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

Convert to centimeters:

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

The computed amplitude is \(6\) cm, which falls within the given range of 6,6.

Answer 2

The total energy (T.E.) in SHM is the sum of kinetic energy (K.E.) and potential energy (P.E.):
\[T.E. = K.E. + P.E.\]
Substitute $K.E. = 0.5 \, \text{J}$ and $P.E. = 0.4 \, \text{J}$:
\[T.E. = 0.5 + 0.4 = 0.9 \, \text{J}.\]
The total energy is also given by:
\[T.E. = \frac{1}{2} m \omega^2 A^2,\]
where:
\[\omega = 2 \pi f \quad \text{and} \quad f = \frac{25}{\pi}.\]
Substitute $\omega$:
\[\omega = 2 \pi \times \frac{25}{\pi} = 50 \, \text{rad/s}.\]
Substitute $T.E. = 0.9 \, \text{J}$, $m = 0.2 \, \text{kg}$, $\omega = 50 \, \text{rad/s}$:
\[0.9 = \frac{1}{2} \times 0.2 \times (50)^2 \times A^2.\]
Simplify:
\[0.9 = 0.1\times 2500 \times A^2 \implies A^2 = \frac{0.9}{2500} = 0.0036.\]
Solve for $A$:
\[A = \sqrt{0.0036} = 0.06 \, \text{m}.\]
Convert to centimeters:
\[A = 6 \, \text{cm}.\]
Thus, the amplitude of oscillation is:
\[A = 6 \, \text{cm}.\]