Question

The total mechanical energy of a spring-mass system in simple harmonic motion is E=12mω2A2. 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 √2E
(d) remain E

Answer 1

The new mechanical energy is E (option D).

The total mechanical energy of the oscillating particle = E = 1/2 mw² A²

m - mass

w - angular velocity of the oscillating particle

A - amplitude of the oscillating particle

Time period of the particle = T = 2π√(m/K)

=> T / 2π = √(m/K) [2π/T = w]

=> w = √(K/m)

Total mechanical energy = 1/2 mw² A²

=> 1/2 m × K/m × A² [w = √(K/m)]

=> 1/2 KA²

When the mass of the particle is doubled keeping the amplitude same , the energy of the particle will remain same because total mechanical energy is independent of the mass of the particle.

The new mechanical energy will remain E.

Answer 2

The total mechanical energy of a spring-mass system in simple harmonic motion is $E=12 m \omega^{2} A^{2}$. 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 remain E.

Explanation:

In a simple harmonic motion, a spring-mass system’s mechanical energy (E) is shown as  

$E=12 m \omega^{2} A^{2}$

where m is body’s mass and ω is angular frequency.

Let the mass be denoted as m1 and the angular frequency be denoted ω1.

New angular frequency ω1 is shown as,

$\omega 1=k ^2 m$          

$(m 1=2 m)$

New energy E1 is shown as,

$E_{1}=12 m_{1} \omega_{1}^{2} A^{2}=12 m \omega^{2} A^{2}$.  

The new mechanical energy as $12 m \omega^{2} A^{2}$.