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

पर्याय

• become 2E

• become E/2

• become \[\sqrt{2}E\]

• remain E

Answer 1

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,

\[E_1 = \frac{1}{2} m_1 \omega_1^2 A^2 \]

\[ = \frac{1}{2}(2m)(\sqrt{\frac{k}{2m}} )^2 A^2 \]

\[ = \frac{1}{2}m \omega^2 A^2 = E\]