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

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:
$A = 6 \, \text{cm}.$

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:
$A = 6 \, \text{cm}.$