A pebble is dropped in a lake, and it produces ripples with a frequency of 0.25 Hz. When should a second pebble be dropped at the same place to produce destructive interference?

Options

A. 0.50 s after the first
B. 1.0 s after the first
C. 2.0 s after the first
D. 0.75 s after the first

Answer

To solve the problem of when to drop the second pebble to produce destructive interference with the ripples created by the first pebble, we can follow these steps:

### Step 1: Understand the Concept of Destructive Interference
Destructive interference occurs when two waves meet and their phase difference is equal to \( \pi \) radians (or 180 degrees). This means that when one wave is at its peak, the other wave is at its trough, effectively canceling each other out.

### Step 2: Determine the Frequency and Period of the Waves
The frequency of the wave produced by the first pebble is given as \( f = 0.25 \, \text{Hz} \). The period \( T \) of the wave can be calculated using the formula:
\[ T = \frac{1}{f} \]
Substituting the frequency:
\[ T = \frac{1}{0.25} = 4 \, \text{seconds} \]

### Step 3: Calculate the Time Delay for Destructive Interference
For destructive interference to occur, the second pebble must be dropped such that the phase difference between the two waves is \( \pi \). This corresponds to a time delay of half the period of the wave:
\[ \Delta t = \frac{T}{2} \]
Substituting the period we calculated:
\[ \Delta t = \frac{4}{2} = 2 \, \text{seconds} \]

### Step 4: Conclusion
The second pebble should be dropped 2 seconds after the first pebble to achieve destructive interference.

### Final Answer
The second pebble should be dropped **2 seconds** after the first pebble.

---