Question

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

• A. $0.15\ \text{m}$
• B. $0.125\ \text{m}$
• C. $0.075\ \text{m}$
• D. $0.1\ \text{m}$

Hint:

To simplify calculations with decimals under exam conditions, convert them into clean fractions! Writing $3.2 = \frac{16}{5}$, $m = \frac{2}{5}$, and $\frac{1}{2}m = \frac{1}{5}$ makes the equation become: $\frac{16}{5} = \frac{1}{5} \times 1024 \times A^2 \implies 16 = 1024 A^2 \implies A^2 = \frac{16}{1024} = \frac{1}{64}$. This completely removes multi-digit decimal division!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The problem describes an object undergoing simple harmonic motion (S.H.M.). We are given the mass, the frequency of oscillation, and the instantaneous values of kinetic and potential energy at a specific displacement. We need to find the total amplitude of oscillation.

Step 2: Key Formula or Approach:
1. Find the total mechanical energy ($E$) of the system by summing the kinetic energy ($K.E.$) and potential energy ($P.E.$): $$E = K.E. + P.E.$$ 2. Relate the total energy to the amplitude ($A$) using the maximum energy formula: $$E = \frac{1}{2} m \omega^2 A^2$$ 3. Find the angular frequency ($\omega$) from the given frequency ($f$): $$\omega = 2\pi f$$

Step 3: Detailed Explanation:
Let's first compute the total energy ($E$) from the given values ($K.E. = 2\ \text{J}$ and $P.E. = 1.2\ \text{J}$): $$E = 2 + 1.2 = 3.2\ \text{J}$$ Now, compute the angular frequency ($\omega$) using the given frequency $f = \frac{16}{\pi}\ \text{Hz}$: $$\omega = 2\pi \left(\frac{16}{\pi}\right) = 32\ \text{rad/s}$$ Substitute total energy $E = 3.2\ \text{J}$, mass $m = 0.4\ \text{kg}$, and $\omega = 32\ \text{rad/s}$ into the total energy equation: $$3.2 = \frac{1}{2} (0.4) (32)^2 A^2$$ $$3.2 = 0.2 \times 1024 \times A^2$$ $$3.2 = 204.8 A^2$$ Isolating $A^2$: $$A^2 = \frac{3.2}{204.8} = \frac{32}{2048} = \frac{1}{64}$$ Taking the positive square root to find the amplitude: $$A = \sqrt{\frac{1}{64}} = \frac{1}{8} = 0.125\ \text{m}$$

Step 4: Final Answer:
The amplitude of oscillation is $0.125\ \text{m}$, which matches option (B).