Read the following passage and answer the accompanying questions - Leaving Cert Physics - Question 11 - 2013

To find the distance to the earthquake's center, we can use the formula:

$s = vt$

where:

• $s$ is the distance,
• $v$ is the speed of the wave (5 km/s), and
• $t$ is the time taken (27 s).

Substituting the values:

$s = 5 ext{ km/s} \times 27 ext{ s} = 135 ext{ km}$

Thus, the station is 135 km away from the earthquake's center.

In the diagram, we show the suspended mass, where the following forces act:

1. Weight (W) acting downwards: $W = mg$
   where $m$ is the mass of the object and $g$ is the acceleration due to gravity (9.8 m/s²).

2. Tension (T) acting upwards from the spring.

At rest, these forces are balanced: $T = W$

To calculate the mass, we use the equation for tension:

$T = mg$
Substituting the known values:

$49 ext{ N} = m \times 9.8 ext{ m/s}^2$

Solving for $m$:

$m = \frac{49}{9.8} = 5 ext{ kg}$

Therefore, the suspended mass is 5 kg.

The frame exhibits simple harmonic motion relative to the suspended mass as it responds to the up and down movement caused by seismic waves.

The acceleration of the ground can be expressed using the formula:

$a = \frac{4\pi^2 s}{T^2}$

where:

• $a$ is the acceleration,
• $s$ is the maximum distance moved,
• $T$ is the periodic time of the wave motion.

Given the period $T = 17$ s and the amplitude $A = 0.8$ cm = 0.008 m, we can calculate the maximum acceleration using:

1. First, convert amplitude to meters, $A = 0.008$ m.
2. Use the formula:

$a_{max} = \frac{4\pi^2 A}{T^2} \Rightarrow a_{max} = \frac{4\pi^2(0.008 m)}{(17 s)^2} \approx 0.0011 \text{ m/s}^2$.

The relative motion between the magnet and the coil induces an electromotive force (emf) in the coil as per Faraday's law of electromagnetic induction. As the magnetic field changes around the coil due to the movement, it generates a voltage across the coil, which can then be measured.