Question 340: Oscillations

Question

A mass $$0.4 \mathrm{~kg}$$ performs S.H.M. with a frequency $$\frac{16}{\pi} \mathrm{Hz}$$. At a certain displacement it has kinetic energy $$2 \mathrm{~J}$$ and potential energy $$1.2 \mathrm{~J}$$. The amplitude of oscillation is

Accepted Answer

To determine the amplitude of the oscillation for the given simple harmonic motion (SHM), we first use some basic principles and formulas related to SHM.
Given:
Mass, $$m = 0.4 \mathrm{~kg}$$
Frequency, $$f = \frac{16}{\pi} \mathrm{Hz}$$
Kinetic energy, $$KE = 2 \mathrm{~J}$$ at a certain displacement
Potential energy, $$PE = 1.2 \mathrm{~J}$$ at the same displacement
First, the total energy in the SHM system is the sum of kinetic energy and potential energy:
$$E_{	ext{total}} = KE + PE = 2 \mathrm{~J} + 1.2 \mathrm{~J} = 3.2 \mathrm{~J}$$
In SHM, the total energy is also given by the formula:
$$E_{	ext{total}} = \frac{1}{2} m \omega^2 A^2$$
where:
$$m$$ is the mass.
$$\omega$$ is the angular frequency.
$$A$$ is the amplitude of oscillation.
First, we need to find the angular frequency $$\omega$$ from the given frequency $$f$$:
$$\omega = 2\pi f$$
Substituting the given frequency, we get:
$$\omega = 2\pi \left(\frac{16}{\pi}ight) = 32 \mathrm{~rad/s}$$
Now, we use the total energy formula to find the amplitude $$A$$:
$$3.2 = \frac{1}{2} 	imes 0.4 	imes 32^2 	imes A^2$$
Simplifying, we get:
$$3.2 = 0.2 	imes 1024 	imes A^2$$
$$3.2 = 204.8 A^2$$
$$A^2 = \frac{3.2}{204.8} = 0.015625$$
$$A = \sqrt{0.015625} = 0.125 \mathrm{~m}$$
Hence, the amplitude of oscillation is $$0.125 \mathrm{~m}$$.
Therefore, the correct option is:
Option B
$$0.125 \mathrm{~m}$$

Answer

$$0.15 \mathrm{~m}$$

Answer

$$0.125 \mathrm{~m}$$

Answer

$$0.075 \mathrm{~m}$$

Answer

$$0.1 \mathrm{~m}$$

Oscillations

A mass $0.4 \mathrm{~kg} performs S.H.M. with a frequency \frac{16}{\pi} \mathrm{Hz}. At a certain displacement it has kinetic energy 2 \mathrm{~J} and potential energy 1.2 \mathrm{~J}$. The amplitude of oscillation is

Choose the correct answer: