Question

The total mechanical energy of a spring-mass system in simple harmonic motion is $E=\frac{1}{2}m{\omega }^2{A}^2.$ 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 E/2
(c) become $\sqrt{2}E$
(d) remain E

Answer 1

(d) remain E
Mechanical energy (E) of a spring-mass system in simple harmonic motion is given by,
$E_{}=\frac{1}{2}m{\omega }^2{A}^2$
where m is mass of body, and
$\omega$ is angular frequency.

Let m₁ be the mass of the other particle and ω₁ be its angular frequency.
New angular frequency ω₁ is given by,
${\omega }_1=\sqrt{\frac{k}{m_1}}=\sqrt{\frac{k}{2m}}(m_1=2m)$

New energy E₁ is given as,
$$\begin{gathered} E_1=\frac{1}{2}{m}_1{\omega }_1^2{A}^2 \\ =\frac{1}{2}\left(2m\right)(\sqrt{\frac{k}{2m}}{)}^2{A}^2 \\ =\frac{1}{2}m{\omega }^2{A}^2=E \end{gathered}$$