QJEEM24Phb3da: JEE Main 2024 Physics - Class 11 JEEM24Phb3da Solution: An object of mass executes simple harmonic motion along axis…

To solve for the amplitude of oscillation, we start by using the properties of simple harmonic motion (SHM). In SHM, the total energy of the system is conserved and is given by the sum of kinetic energy (KE) and potential energy (PE).

Given:

• Mass $m = 0.2 \ \mathrm{kg}$
• Frequency $f = \left(\frac{25}{\pi}\right) \ \mathrm{Hz}$
• Position $x = 0.04 \ \mathrm{m}$
• KE at $x = 0.04 \ \mathrm{m}$ is $0.5 \ \mathrm{J}$
• PE at $x = 0.04 \ \mathrm{m}$ is $0.4 \ \mathrm{J}$

The total mechanical energy (E) of the SHM system can be found by summing the given kinetic and potential energies:

$E = KE + PE = 0.5 \ \mathrm{J} + 0.4 \ \mathrm{J} = 0.9 \ \mathrm{J}$

For simple harmonic motion, the total energy (E) is also related to the amplitude (A) by the following formula:

$E = \frac{1}{2} k A^2$

where $k$ is the spring constant. First, we need to find the angular frequency $\omega$ :

$\omega = 2 \pi f = 2 \pi \left(\frac{25}{\pi}\right) \ \mathrm{Hz} = 50 \ \mathrm{rad/s}$

The spring constant $k$ can be calculated using the relationship between $m$ , $\omega$ , and $k$ :

$\omega = \sqrt{\frac{k}{m}} \Rightarrow k = m \omega^2 = 0.2 \times (50)^2 = 500 \ \mathrm{N/m}$

Now, substituting $k$ back into the energy equation, we solve for the amplitude $A$ :

$0.9 = \frac{1}{2} \times 500 \times A^2 \Rightarrow A^2 = \frac{0.9 \times 2}{500} \Rightarrow A^2 = \frac{1.8}{500} \Rightarrow A^2 = 0.0036 \Rightarrow A = \sqrt{0.0036} = 0.06 \ \mathrm{m}$

Converting $A$ from meters to centimeters:

$A = 0.06 \ \mathrm{m} \times 100 = 6 \ \mathrm{cm}$

Thus, the amplitude of oscillation is $6 \ \mathrm{cm}$ .