A wire having a linear mass density 5.0 x 10-3 kg/m is stretched betwe...
A wire having a linear mass density 5.0 x 10-3 kg/m is stretched betwe...
To find the length of the wire, we can use the formula for the fundamental frequency of a stretched wire:
f = (1/2L) * √(T/μ)
where:
f is the frequency of vibration,
L is the length of the wire,
T is the tension in the wire, and
μ is the linear mass density of the wire.
We are given the following information:
Tension, T = 450 N
Linear mass density, μ = 5.0 x 10^-3 kg/m
Fundamental frequency, f = 400 Hz
Let's substitute these values into the formula and solve for L:
400 = (1/2L) * √(450 / 5.0 x 10^-3)
Simplifying the equation:
400 = (1/2L) * √(9 x 10^4 / 5.0 x 10^-3)
400 = (1/2L) * √(9 x 10^4 x 10^3 / 5.0)
400 = (1/2L) * √(9 x 10^7 / 5.0)
400 = (1/2L) * √(1.8 x 10^7)
Now, to find the length of the wire, we need to solve for L:
L = (1/2) * √(1.8 x 10^7) / 400
L = (0.5) * √(1.8 x 10^7) / 400
L = 0.5 * 4242.64 / 400
L = 2121.32 / 400
L ≈ 5.3033 m
Since the problem asks for the next higher frequency at which the wire resonates, we need to find the length of the wire for a frequency of 500 Hz.
Let's substitute the new frequency into the formula and solve for L:
500 = (1/2L) * √(450 / 5.0 x 10^-3)
Simplifying the equation:
500 = (1/2L) * √(9 x 10^4 / 5.0 x 10^-3)
500 = (1/2L) * √(9 x 10^4 x 10^3 / 5.0)
500 = (1/2L) * √(9 x 10^7 / 5.0)
500 = (1/2L) * √(1.8 x 10^7)
Now, solving for L:
L = (1/2) * √(1.8 x 10^7) / 500
L = (0.5) * √(1.8 x 10^7) / 500
L = 0.5 * 4242.64 / 500
L = 2121.32 / 500
L ≈ 4.2426 m
Since the next higher frequency occurs at 500 Hz, the length of the wire is approximately 4.2426 m. Therefore, the correct answer is option D) 1.5 m.
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