0.5 kg mass is attached to a spring with a spring constant of 100 n/m if the mass is moving with a velocity of 2 m/s and is displaced by 0.3from it's equilibrium position find T.E of S.H.M

1 answer

To find the total energy (T.E.) of the simple harmonic motion (S.H.M.) of a mass attached to a spring, we need to consider both the potential energy (P.E.) stored in the spring and the kinetic energy (K.E.) of the mass.

  1. Kinetic Energy (K.E.): The kinetic energy of the mass can be calculated using the formula: \[ K.E. = \frac{1}{2} m v^2 \] where

    • \(m = 0.5 , \text{kg}\) (mass),
    • \(v = 2 , \text{m/s}\) (velocity).

    Plugging in the values: \[ K.E. = \frac{1}{2} \times 0.5 \times (2)^2 = \frac{1}{2} \times 0.5 \times 4 = 1 , \text{J} \]

  2. Potential Energy (P.E.): The potential energy stored in the spring when it is displaced by a distance \(x\) from its equilibrium position is given by: \[ P.E. = \frac{1}{2} k x^2 \] where

    • \(k = 100 , \text{N/m}\) (spring constant),
    • \(x = 0.3 , \text{m}\) (displacement).

    Plugging in the values: \[ P.E. = \frac{1}{2} \times 100 \times (0.3)^2 = \frac{1}{2} \times 100 \times 0.09 = \frac{1}{2} \times 9 = 4.5 , \text{J} \]

  3. Total Energy (T.E.): The total energy in the system is the sum of the kinetic and potential energies: \[ T.E. = K.E. + P.E. \] Plugging in the values: \[ T.E. = 1 , \text{J} + 4.5 , \text{J} = 5.5 , \text{J} \]

Thus, the total energy of the S.H.M. is \(5.5 , \text{J}\).