Question 374: Oscillations

Question

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

Accepted Answer

To find the amplitude of oscillation, let's start by considering the given data and the formulae related to simple harmonic motion (SHM).
The total mechanical energy of a system in SHM is constant and is the sum of kinetic energy (KE) and potential energy (PE) at any point in its motion. We are given:
Kinetic Energy, $$ KE = 0.5 \, 	ext{J} $$
Potential Energy, $$ PE = 0.4 \, 	ext{J} $$
Therefore, the total energy (E) is:
$$ E = KE + PE = 0.5 + 0.4 = 0.9 \, 	ext{J} $$
In SHM, the total mechanical energy can also be expressed as:
$$ E = \frac{1}{2} k A^2 $$
where $ k $ is the spring constant and $ A $ is the amplitude.
We can also express kinetic and potential energy in SHM in terms of amplitude $ A $, position $ x $, and angular frequency $ \omega $:
Kinetic Energy: $$ KE = \frac{1}{2} m \omega^2 (A^2 - x^2) $$
Potential Energy: $$ PE = \frac{1}{2} m \omega^2 x^2 $$
Given mass $ m = 0.2 \, 	ext{kg} $, position $ x = 0.04 \, 	ext{m} $.
The angular frequency $ \omega $ is related to the frequency $ f $ by:
$$ \omega = 2\pi f $$
Given frequency:
$$ f = \frac{25}{\pi} \, 	ext{Hz} $$
Thus,
$$ \omega = 2\pi \left(\frac{25}{\pi}ight) = 50 \, 	ext{rad/s} $$
Using the expression for the total energy in SHM:
$$ \frac{1}{2} k A^2 = \frac{1}{2} m \omega^2 A^2 $$
Since the total energy $ E = 0.9 \, 	ext{J} $, it follows:
$$ 0.9 = \frac{1}{2} \cdot 0.2 \cdot (50)^2 \cdot A^2 $$
Solving for $ A^2 $:
$$ 0.9 = \frac{1}{2} \cdot 0.2 \cdot 2500 \cdot A^2 $$
$$ 0.9 = \frac{1}{2} \cdot 500 \cdot A^2 $$
$$ 0.9 = 250 \cdot A^2 $$
$$ A^2 = \frac{0.9}{250} $$
$$ A^2 = 0.0036 $$
$$ A = \sqrt{0.0036} $$
$$ A = 0.06 \, 	ext{m} $$
The amplitude of the oscillations is
0.06 m
.

Oscillations

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

Integer / Numeric Answer Type