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The G string on a violin has a fundamental frequency of 196 Hz. It is 30.0 cm long. While this string is sounding, a nearby violinist effectively shortens (by sliding her finger down the string) the G string on her identical violin until a beat frequency of 3.5 Hz is heard between the two strings. When this occurs, how far down the string did she slide her finger

User Entesar
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2 Answers

7 votes

Final answer:

The violinist has slid her finger down by approximately 0.53 cm on the G string to create the 3.5 Hz beat frequency when compared with another G string of the same violin vibrating at 196 Hz.

Step-by-step explanation:

To find out how far down the string the violinist slid her finger to create a beat frequency of 3.5 Hz, we first need to determine the frequency of the second vibrating string.

Knowing that one violin string is already vibrating at a fundamental frequency of 196 Hz, we can calculate the frequency of the second string by adding or subtracting the beat frequency, as beat frequency is the absolute difference between the two frequencies.

Thus, the frequency of the second string could be either f2 = 196 Hz + 3.5 Hz or f2 = 196 Hz - 3.5 Hz. Let's choose the higher frequency for this instance, so we have f2 = 199.5 Hz.

Next, we use the formula for the fundamental frequency of a string, which is f1 = v/(2L), where f1 is the fundamental frequency, v is the wave speed on the string, and L is the length of the string.

Since the string lengths are initially identical and the wave speeds are the same on both strings, we can express the second length, L2, as L0 - x, where L0 is the initial length and x is the length the violinist's finger has covered.

Applying the fact that the frequencies are inversely proportional to the lengths (assuming string tension and mass per unit length remain constant), we can set up the following equation: f1/f2 = (L0 - x) / L0.

Now, we calculate the length the string has been shortened: 196 Hz / 199.5 Hz = (30.0 cm - x) / 30.0 cm. Solving for x yields x = 30.0 cm (1 - 196/199.5), which results in x = 0.53 cm. Therefore, the violinist has slid her finger down the string by 0.53 cm to create the 3.5 Hz beat frequency.

User Amer Qarabsa
by
4.6k points
3 votes

Answer:

L = 29.47 cm

Step-by-step explanation:

wavelength = twice the string length = (wave velocity on string)/(frequency)

Thus;

2(L_o) = v/(f_o)

v = 2(L_o)(f_o) - - - eq1

wave velocity on string = √[(tension)/(mass per unit length)]

If we assume that the 2 violins have matching string gauges and they're tuned exactly the same, and the necks of the violins are the same length, the tensions needed to get the same frequency on open G strings would be the same. Hence, we can deduce that the wave velocities on the 2 strings would also be the same.

The note on one violin has to differ from the note on the other violin by the beat frequency. A shorter string produces a higher pitch (higher frequency). Thus,

v = 2L(f_o + f_b) - - - - eq2

Equating equation 1 and 2 to obtain;

2(L_o)(f_o) = 2L(f_o + f_b)

(Lo)(fo) = L(fo + fb)

(30 cm)(196 Hz) = (L)(196Hz + 3.5Hz)

5880 = 199.5L

L = 5880/199.5

L = 29.47 cm

User Pingui
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