The total mechanical energy of a spring-mass system in 1 simple h… - Vidyadip

MCQ

The total mechanical energy of a spring$-$mass system in $1$ simple harmonic motion is $\text{E}=\frac{1}{2}\text{m}\omega^2\text{A}^2.$ Suppose the oscillating particle is replaced by another particle of double the mass while the amplitude $A$ remains the same. The new mechanical energy will:

A
Become $2E$
B
Become $\frac{\text{E}}{2}$
C
Become $\sqrt{2\text{E}}$
✓
Remain $E.$

✓

Answer

Correct option: D.

Remain $E.$

Mechanical energy $(E)$ of a spring$-$mass system in simple harmonic motion is given by,
$\text{E}=\frac{1}{2}\text{m}\omega^2\text{A}^2$
where $m$ is mass of body, and $\omega$ is angular frequency.
Let $m_1$ be the mass of the other particle and $\omega_1$ be its angular frequency.
New angular frequency $\omega_1$ is given by,
$\omega_1=\sqrt{\frac{\text{k}}{\text{m}_1}}=\sqrt{\frac{\text{k}}{2\text{m}}}\big(\text{m}_1=2\text{m}\big)$
New energy $E_1$ is given as,
$\text{E}_1=\frac{1}{2}\text{m}_1\omega_1^2\text{A}^2$
$=\frac{1}{2}\big(2\text{m}\big)\Big(\sqrt{\frac{\text{k}}{2\text{m}}}\Big)^2\text{A}^2$
$=\frac{1}{2}\text{m}\omega^2\text{A}^2=\text{E}$

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