copper Rod of length l=50 cm is clamped at its midpoint .find the number of natural longitudinal oscillations of the Rod In the frequency range from 20 to 50kHz, what are those frequencies equal to
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To determine the number of natural longitudinal oscillations of a copper rod clamped at its midpoint, we need to consider the fundamental frequency of vibration.
The fundamental frequency can be obtained from the formula:
f = (v / (2L))
where:
f represents the frequency,
v is the speed of sound through the material (copper in this case),
L is the length of the rod.
Given that the length of the copper rod is l = 50 cm = 0.5 m, we can calculate the fundamental frequency using the above formula.
Now, let's find the speed of sound through copper. The speed of sound in any material depends on its density and bulk modulus. For copper, the average speed of sound is approximately v = 3800 m/s.
Using the given values, we can substitute them into the formula to find the fundamental frequency:
f = (3800 / (2 * 0.5))
f = 3800 / 1
f = 3800 Hz
Therefore, the fundamental frequency of oscillation for the copper rod clamped at its midpoint is 3800 Hz (or 3.8 kHz).
To determine the number of natural longitudinal oscillations in the frequency range from 20 to 50 kHz, we can divide the frequency range by the fundamental frequency:
Number of oscillations = (50,000 - 20,000) / 3800
Number of oscillations = 30,000 / 3800
Number of oscillations = 7.89
Since we can't have a fraction of an oscillation, the number of natural longitudinal oscillations in the frequency range from 20 to 50 kHz would be 7 oscillations.