State Hooke's law - Leaving Cert Physics - Question a - 2014

To find the new length of the spring after the mass is attached:

1. Calculate the force due to the mass:

   $F = mg = 0.02 ext{ kg} imes 9.8 ext{ m/s}^2 = 0.196 ext{ N}$

2. Use Hooke's law to find the extension ($x$):

   $x = \frac{F}{k} = \frac{0.196}{12} = 0.0163 ext{ m} = 16.3 ext{ mm}$

3. Add the extension to the original length of the spring:

   New Length = Original Length + Extension = 25 ext{ mm} + 16.3 ext{ mm} = 41.3 ext{ mm}.

The velocity-time graph for the object's simple harmonic motion should:

1. Have the horizontal axis representing time and the vertical axis representing velocity.
2. Start at zero velocity (when the object is at maximum displacement).
3. Show a sinusoidal curve oscillating above and below the time axis, indicating positive and negative velocities.
4. Be periodic in nature, demonstrating that the object oscillates back and forth with a constant amplitude.

To calculate the period ($T$) of the oscillation:

1. Apply the formula for the period of a spring-mass system:

   $T = 2\pi \sqrt{\frac{m}{k}}$

2. Substitute the values:

   $T = 2\pi \sqrt{\frac{0.02 ext{ kg}}{12 ext{ N/m}}} = 2\pi \sqrt{\frac{0.02}{12}} \approx 0.256 ext{ s}.$

Thus, the period of oscillation is approximately 0.256 seconds.