A mass at the end of a spring obeys Hooke's law - Leaving Cert Physics - Question 6 - 2016

Hooke's law states that the force exerted by a spring is proportional to the displacement from its rest position. Mathematically, this can be expressed as:

$F = -ks$

where $F$ is the restoring force, $k$ is the spring constant, and $s$ is the displacement.

To demonstrate that a mass obeying Hooke's law executes simple harmonic motion, we start from the equation derived from Hooke's law:

The acceleration of the mass can be expressed as: $a = \frac{F}{m} = \frac{-ks}{m}$ This shows that the acceleration is directly proportional to the negative displacement, confirming that the motion is simple harmonic.

To find the length of the pendulum, we use the formula for the period of a simple pendulum:

$T = 2\pi \sqrt{\frac{L}{g}}$ Given that the period $T = 2$ s and $g = 9.8$ m/s², we rearrange for $L$:

$L = \frac{gT^2}{4\pi^2} = \frac{9.8 \times (2)^2}{4\pi^2} = 0.99 ext{ m}$

The maximum angular displacement can be calculated by finding the relationship between the arc length $s$, the radius (length of pendulum $L = 0.99$ m), and the angle $heta$:

$s = L \theta \implies \theta = \frac{s}{L} = \frac{0.18}{0.99} = 0.045 ext{ m} \implies \theta = 0.045 ext{ radians}$

The diagram should include:

1. The weight of the bob acting downward (W).
2. The tension in the string acting at an angle to the vertical. At maximum displacement, this is closer to horizontal. Label the angle with respect to the vertical at this point.

At maximum displacement, the restoring force ($F$) can be calculated by analyzing the vertical components. The formula is given by:

$F = W \sin(\theta)$ Given that the weight $W = 3.5$ N and $heta$ is derived previously, if we consider $heta$ a small value, for our calculations let $F \approx 0.16$ N.

The bob achieves its greatest angular velocity at the lowest point of its swing (the lowest point in the arc). This is where all potential energy has been converted into kinetic energy, maximizing the speed and hence angular velocity.

Let $T_{surface}$ be the period at the surface. A 2% increase gives us:

$T_{new} = 1.02 T_{surface}$ Using: $T = 2\pi \sqrt{\frac{L}{g}}$ We can set the equations up with the new height: $T_{new}^2 = \frac{4\pi^2 L}{g_{new}}$ Using $g_{new} = g(\frac{R}{R+h})^2$, solve for the height $h$ which gives approximately 127.4 km above the surface.