he largest tension that can be sustained by a stretched string of linear mass density μ, even in principle, is given by τ = μc2, where c is the speed of light in vacuum. (This is an enormous value. The breaking tensions of all ordinary materials are about 12 orders of magnitude less than this.)

a) What is the speed of a traveling wave on a string under such tension?

b) If a 1.170-m-long guitar string, stretched between anchored ends, were made of this hypothetical material, what frequency would its first harmonic have?

c) If that guitar string were plucked at its midpoint and given a displacement of 1.89 mm there to produce the fundamental frequency, what would be the maximum speed attained by the midpoint of the string?

a)The speed of sound

a) To find the speed of a traveling wave on a string under the given tension, we can use the wave equation v = sqrt(T/μ), where v is the speed of the wave, T is the tension, and μ is the linear mass density of the string.

Given the tension τ = μc^2, we can substitute this value into the wave equation:

v = sqrt(τ/μ) = sqrt((μc^2)/μ) = sqrt(c^2) = c

Therefore, the speed of a traveling wave on a string under such tension is equal to the speed of light in vacuum, c.

b) To find the frequency of the first harmonic of a guitar string made of the hypothetical material, we can use the formula f = v/(2L), where f is the frequency, v is the speed of the wave, and L is the length of the string.

Given the length of the guitar string, L = 1.170 m, and the speed of the wave, v = c, we can substitute these values into the formula:

f = c/(2L) = (3.00 x 10^8 m/s)/(2 x 1.170 m) ≈ 1.28 x 10^8 Hz

Therefore, the first harmonic of the guitar string would have a frequency of approximately 1.28 x 10^8 Hz.

c) To find the maximum speed attained by the midpoint of the guitar string when plucked at its midpoint and given a displacement, we can use the formula vmax = ωA, where vmax is the maximum speed, ω is the angular frequency, and A is the amplitude of the displacement.

For the fundamental frequency, the angular frequency is given by ω = 2πf, where f is the frequency. The amplitude can be converted to meters, as 1.89 mm = 0.00189 m.

Given the fundamental frequency from part b, f = 1.28 x 10^8 Hz, and the amplitude A = 0.00189 m, we can substitute these values into the formula:

vmax = (2πf)A = (2π x 1.28 x 10^8 Hz) x 0.00189 m ≈ 1.63 x 10^6 m/s

Therefore, the maximum speed attained by the midpoint of the guitar string would be approximately 1.63 x 10^6 m/s.