Question

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

Answer 1

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 \(5.5 \, \text{J}\).