Animated Solution for Physics - Waves: A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7×103 kg/m3 and its Young's modulus is 9.27×1010 Pa. What will be the fundamental frequency of the longitudinal vibrations?
Select Answer:
Visualized Solution
PhysicalSetup
Rod of length L=60 cm clamped at the middle.
It is set into longitudinal vibrations.
FundamentalMode
Clamped middle →Node (N)
Free ends →Antinodes (A)
L=2λ⟹λ=2L
SpeedofLongitudinalWave
v=ρY
FundamentalFrequency
f=λv=2L1ρY
Substitution
f=2(0.6)12.7×1039.27×1010
CalculatingWaveSpeed
v=34.33×106≈5860 m/s
FinalFrequency
f=1.25860≈4883 Hz
f≈5 kHz
TheWayForward
What if the rod was clamped at 4L?
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The Sigma Insight: Standing Waves in Strings and Organ Pipes
Solution Diagram
The Symphony of a Granite Rod
Unraveling Longitudinal Vibrations
Imagine a solid granite rod, exactly 60 cm long, firmly clamped right at its center. When you strike the end of this rod with a hammer, it doesn't just bend; it compresses and expands along its length. This creates a longitudinal standing wave.
But how do we find the fundamental frequency of this invisible vibration? Let's break down the physics step by step.
The Anatomy of the Wave
Because the rod is clamped at the middle, that specific point is physically restricted from moving. In the language of waves, a point of zero displacement is called a Node (N).
Conversely, the two free ends of the rod have the maximum freedom to oscillate back and forth. These points of maximum displacement are called Antinodes (A).
For the rod to vibrate in its simplest, lowest-frequency pattern (the fundamental mode), it must form the simplest possible standing wave that fits these boundary conditions: an antinode at one end, a node in the middle, and an antinode at the other end.
The distance between a node and an adjacent antinode is always 4λ. Since our rod has two such segments (from the center to each end), the total length L of the rod is:
L=4λ+4λ=2λ
This tells us that the wavelength of the fundamental mode is exactly twice the length of the rod:
λ=2L
The Speed of Sound in Granite
Before we can find the frequency, we need to know how fast the wave travels through the granite. The speed of a longitudinal wave in a solid rod depends on two intrinsic properties of the material: its elasticity (Young's modulus, Y) and its inertia (density, ρ). The formula is beautifully simple:
v=ρY
Let's plug in the given values for our granite rod:
- Young's modulus, Y=9.27×1010 Pa
- Density, ρ=2.7×103 kg/m3
v=2.7×1039.27×1010=34.33×106≈5859 m/s
This means the wave zips through the granite at nearly 6 km/s!
The Grand Finale
Calculating Frequency
Now that we have both the wave speed and the wavelength, finding the frequency is straightforward. We use the universal wave equation, v=fλ, rearranged for frequency:
f=λv=2Lv
Substitute the speed we just calculated and the length of the rod (L=60 cm=0.6 m):
f=2×0.65859=1.25859≈4882.5 Hz
Rounding this off to the nearest given option, we get:
f≈5000 Hz=5 kHz
And there we have it! The granite rod sings a high-pitched note of 5 kHz when struck.