Explanation: "Imagine two sets of waves that have the same speed. If one set has a longer wavelength, it will have a lower frequency (more time between waves). If the other set has a shorter wavelength, it will have a higher frequency (less time between waves). Light moves even faster AND has shorter wavelengths."

Why it's not C: "The number of complete wavelengths in a given unit of time is called frequency (f). As a wavelength increases in size, its frequency and energy (E) decrease. From these equations you may realize that as the frequency increases, the wavelength gets shorter. As the frequency decreases, the wavelength gets longer."

Why it's not B: "The frequency does not change as the sound wave moves from one medium to another. Since the speed changes and the frequency does not, the wavelength must change."

Why it's not A: "Do loud sounds travel faster than soft sounds? No. Both travel at the same speed The speed depends on the medium it passes through. Louder sounds are simply sound waves with higher amplitude traveling at the same speed."

6 0

3 years ago

A string with a mass density of 3 * 10^-3 kg/m is under a tension of 380 N and is fixed at both ends. One of its resonance frequ

Delvig [45]

Answer:

(a) the fundamental frequency of this string is 65 Hz

(b) the harmonics of the given frequencies are third and fourth respectively.

Explanation:

Given;

mass density of the string, μ = 3 x 10⁻³ kg/m

tension of the string, T = 380 N

resonating frequencies, 195 Hz and 260 N

For the given resonant frequencies;

$195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } ---(1)\\\\260 = \frac{n+1}{2l} \sqrt{\frac{T}{\mu} } ---(2)\\\\divide \ (2) \ by (1)\\\\\frac{260}{195} = \frac{n+1 }{n} \\\\260n = 195(n+1)\\\\260 n = 195 n + 195\\\\260n - 195n = 195\\\\65n = 195\\\\n = \frac{195}{65} \\\\n = 3$

$195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } \\\\195 = \frac{3}{2l}\sqrt{\frac{380}{3\times 10^{-3}} } \\\\l = \frac{3}{2\times 195}\sqrt{\frac{380}{3\times 10^{-3}} }\\\\l = 2.74 \ m$