A wire is stretched between two points, 1 m apart. If the speed of the wave generated on plucking the wire is 200 ms\(^{-1}\), what is the minimum frequency which will resonate with the wire?

Joshua322

2 months ago

It looks like myschool.ng made another mistake by assuming the wavelength (
) is equal to the length of the wire (1m). This is a common trap for students aiming for a 300+ score in JAMB!
Here is the "dummie level" breakdown of why they are wrong:
1. The "Wavy Line" Logic
Look at your image again. Specifically, look at one full "S" shape of the wavy red line.
When a string is plucked, the simplest wave it can make is only half of that "S" shape (from the top of the curve to the bottom).
Because the string is fixed at both ends, it creates a "loop."
2. The Relationship Rule
The String Length (
): 1 meter.
The Rule: A string vibrating at its minimum (fundamental) frequency only fits half a wave in that 1 meter.
The Math: If half a wave is 1 meter, then a full wave (
) must be 2 meters.
3. Comparing the Calculations
Step My Calculation (Correct) myschool.ng (Incorrect)
Wavelength (
) 2 m (Double the string) 1 m (Same as string)
Formula



Calculation

Result 100 Hz 200 Hz
Summary for your JAMB 300+ Goal:
Minimum Frequency (Fundamental):

First Overtone:

(This is the only time wavelength equals length!)
Second Overtone: