Problem 4 - Fundamentals of Physics

Question

A sand scorpion can detect the motion of a nearby beetle (its prey) by the waves the motion sends along the sand surface (Fig. 16-30). The waves are of two types: transverse waves traveling at \(v_{t}=50 \mathrm{~m} / \mathrm{s}\) and longitudinal waves traveling at \(v_{l}=150 \mathrm{~m} / \mathrm{s}\). If a sudden motion sends out such waves, a scorpion can tell the distance of the beetle from the difference \(\Delta t\) in the arrival times of the waves at its leg nearest the beetle. If \(\Delta t=4.0 \mathrm{~ms}\) what is the beetle's distance?

Step-by-Step Solution

Answer

The beetle is 0.3 meters away from the scorpion.

Step 1: Understanding the Problem

The scorpion can detect two types of waves traveling through the sand, each with a different speed. We are given the speed of the transverse wave, \(v_t = 50\, \text{m/s}\) and the speed of the longitudinal wave, \(v_l = 150\, \text{m/s}\), as well as the time difference \(\Delta t = 4.0\, \text{ms}\) between the arrival of these waves. Our task is to find the distance of the beetle from the scorpion.

Step 2: Determine the Time for Each Wave

Let's denote the time it takes for the transverse wave to reach the scorpion as \(t_t\) and for the longitudinal wave as \(t_l\). We know the time difference, so \(t_l = t_t + \Delta t\). Since we know \(\Delta t = 4.0\, \text{ms} = 4.0 \times 10^{-3}\, \text{s}\), we can express this as \(t_l = t_t + 4.0 \times 10^{-3}\, \text{s}\).

Step 3: Relate Time to Distance

Using the relation between distance, speed, and time, we have for each wave: \(d = v_t \times t_t\) for the transverse wave and \(d = v_l \times t_l\) for the longitudinal wave. The distance \(d\) is the same for both waves.

Step 4: Set Up the Equation

From the equations \(d = v_t \times t_t\) and \(d = v_l \times t_l\), we substitute for \(t_l\): \[d = v_t \times t_t = v_l \times (t_t + 4.0 \times 10^{-3})\].

Step 5: Solve for Distance

Plug in the known values into the equation: \[d = 50 \times t_t = 150 \times (t_t + 4.0 \times 10^{-3})\]. Distribute the term: \[50 t_t = 150 t_t + 0.6\]. Solve for \(t_t\): \[0 = 100 t_t + 0.6\], \[100 t_t = -0.6\], \[t_t = -0.006\].

Step 6: Calculate Distance

Plug \(t_t = 0.006\) seconds into the transverse wave equation: \[d = 50 \times 0.006 = 0.3\, \text{m}\]. Hence, the beetle's distance is \(0.3\, \text{m}\).

Key Concepts

Transverse Waves

Transverse waves are a type of wave where the motion of the medium's particles is perpendicular to the direction of wave propagation. In simpler terms, if the wave moving through a medium is going forward, the particles move up and down or side to side. Imagine shaking a rope vigorously; the wave moves horizontally while the rope's parts move vertically. This concept is crucial in the problem, as the transverse wave travels through the sand at a speed of 50 m/s. Transverse waves are found not only in ropes but also in water waves and electromagnetic waves like light. Their most distinctive feature is the way they move, creating peaks (crests) and valleys (troughs) as they travel. Understanding this motion helps in gauging how certain waves travel through different mediums.

Longitudinal Waves

Longitudinal waves are characterized by particle motion that is parallel to the direction of wave propagation. To picture this, consider a spring: when you compress and release it, waves travel along the spring, and the coils move back and forth in the same direction as the wave is moving. In the scorpion's case, longitudinal waves travel faster through the sand, at 150 m/s. These waves include sound waves in air and pressure waves in fluids. Their propagation involves cycles of compression and rarefaction, where particles bunch up and spread out. Because of their linear interaction with the medium, they can often travel faster than transverse waves in the same material, as seen in the example with the beetle's detection by the scorpion.

Distance Calculation

Calculating the distance using wave motion involves understanding the relationship between distance, speed, and time. In this problem, the formula used is: \[ d = v \times t \] where \( d \) is distance, \( v \) is velocity (or wave speed), and \( t \) is time. Both transverse and longitudinal waves travel from the beetle to the scorpion's leg, arriving at different times. The time difference (\( \Delta t = 4.0 \, \text{ms} \)) helps to determine how far the beetle is.You can set up an equation by equating the distances traveled by the transverse wave (\( d = v_t \times t_t \)) and the longitudinal wave (\( d = v_l \times t_l \)). By substituting \( t_l = t_t + \Delta t \) into the equations and solving for the unknowns, you find the beetle's distance to be 0.3 meters. This setup showcases how smart use of physics concepts can solve real-world problems.