Desert Beetle Pulses: Determining Prey Distance

This problem involves understanding the propagation of different types of waves through a medium and using their speeds and arrival time differences to calculate the distance to the source. In this scenario, a beetle's movement generates two distinct types of pulses in the desert sand: fast longitudinal pulses and slower transverse pulses. A sand scorpion, detecting these pulses, uses the time difference between their arrivals to pinpoint the location of its prey.

Pulses and Wave Speeds in Sand

When the beetle moves, it creates vibrations that travel through the sand. These vibrations manifest as two types of waves:

• Longitudinal Pulses: These are waves where the particles of the medium oscillate parallel to the direction of wave propagation. They are typically faster in solids. Their speed is given as \(v_L = 150 \text{ ms}^{-1}\).
• Transverse Pulses: These are waves where the particles of the medium oscillate perpendicular to the direction of wave propagation. They are typically slower in solids compared to longitudinal waves. Their speed is given as \(v_S = 50 \text{ ms}^{-1}\).

The sand scorpion detects these pulses. Since the longitudinal pulses travel faster, they will reach the scorpion before the transverse pulses do, even though both originate from the same event (the beetle's motion) at the same distance.

Time Difference and Distance Calculation

Let's denote the unknown distance to the prey (the beetle) as \(d\).

The time taken for the fast longitudinal pulse to travel distance \(d\) is:

\[t_L = \frac{d}{v_L}\]

The time taken for the slower transverse pulse to travel the same distance \(d\) is:

\[t_S = \frac{d}{v_S}\]

According to the problem, the scorpion intercepts the faster longitudinal pulses \(4.0 \text{ ms}\) earlier than the slower transverse pulses. This means the difference in their arrival times is \(4.0 \text{ ms}\). Since the transverse pulse is slower, it takes longer to arrive.

Therefore, the time difference can be expressed as:

\[\Delta t = t_S - t_L\]

We are given \(\Delta t = 4.0 \text{ ms} = 4.0 \times 10^{-3} \text{ s}\).

Substitute the expressions for \(t_S\) and \(t_L\) into the equation:

\[\Delta t = \frac{d}{v_S} - \frac{d}{v_L}\]

Now, we can factor out \(d\):

\[\Delta t = d \left( \frac{1}{v_S} - \frac{1}{v_L} \right)\]

To combine the terms in the parenthesis, find a common denominator:

\[\Delta t = d \left( \frac{v_L - v_S}{v_L v_S} \right)\]

Finally, to solve for \(d\), rearrange the equation:

\[d = \Delta t \left( \frac{v_L v_S}{v_L - v_S} \right)\]

Numerical Calculation of Prey Distance

Let's plug in the given values:

• Longitudinal pulse speed, \(v_L = 150 \text{ ms}^{-1}\)
• Transverse pulse speed, \(v_S = 50 \text{ ms}^{-1}\)
• Time difference, \(\Delta t = 4.0 \times 10^{-3} \text{ s}\)

Substitute these values into the derived formula for \(d\):

\[d = (4.0 \times 10^{-3} \text{ s}) \left( \frac{(150 \text{ ms}^{-1}) (50 \text{ ms}^{-1})}{(150 \text{ ms}^{-1}) - (50 \text{ ms}^{-1})} \right)\]

First, calculate the numerator and denominator within the fraction:

• Numerator: \(150 \times 50 = 7500\) \(\text{m}^2\text{s}^{-2}\)
• Denominator: \(150 - 50 = 100\) \(\text{ms}^{-1}\)

Now, substitute these back:

\[d = (4.0 \times 10^{-3}) \left( \frac{7500}{100} \right)\]

Simplify the fraction:

\[d = (4.0 \times 10^{-3}) (75)\]

Perform the multiplication:

\[d = 300 \times 10^{-3} \text{ m}\]

\[d = 0.300 \text{ m}\]

To convert meters to centimeters, multiply by 100:

\[d = 0.300 \text{ m} \times 100 \text{ cm/m}\]

\[d = 30 \text{ cm}\]

Therefore, the prey is located at a distance of \(30 \text{ cm}\) from the scorpion.

The information about the scorpion having eight legs spread in a circle of 5 cm diameter is additional context and does not directly affect the calculation of the distance to the prey based on the time difference of the pulses traveling along the sand's surface. It likely serves to illustrate the scorpion's ability to sense vibrations.