Question

PART ONE
A cello string vibrates in its fundamental
mode with a frequency of 239 1/s. The vibrating segment is 76.2 cm long and has a
mass of 0.48 g.
Find the tension in the string.
Answer in units of N.

PART TWO (REQUIRES PICTURE)
Find the frequency of the string when it
vibrates in six segments.
Answer in units of 1/s.

Answer 1

1) 83.6 N

2) 1434 Hz

Explanation:

1)

The fundamental frequency of the standing waves on a string is given by the equation

$f_1=\frac{1}{2L}\sqrt{\frac{T}{\mu}}$

where:

L is the length of the string

T is the tension in the string

$\mu$ is the mass per unit length, which can be rewritten as

$\mu=\frac{m}{L}$, where m is the mass of the string

For the string in this problem, we have:

L = 76.2 cm = 0.762 m is the length of the string

$m=0.48 g = 0.48\cdot 10^{-3} kg$ is the mass

$f_1=239 Hz$ is the fundamental frequency

Re-arranging the equation for T, we find the tension in the string:

$T=(2Lf_1)^2 \mu = (2Lf_1)^2 \frac{m}{L}=(2(0.762)(239))^2 \frac{0.48\cdot 10^{-3}}{0.762}=83.6 N$

2)

In this problem, we see that the string vibrates in six segments.

A standing wave on a string has different modes of vibrations, characterized by different frequencies:

- In the  fundamental mode ($f_1$), the string vibrates in one segment

- In the second harmonic ($f_2$), the string vibrates in 2 segments

... and so on

The frequency of the nth-harmonic is given by

$f_n = nf_1$

where $f_1$ is the fundamental frequency.

In this problem, we know that

$f_1=239 Hz$

And the string is vibrating in 6 modes, so we are finding the frequency of the 6th harmonic, which is

$f_6 = 6f_1 = 6(239)=1434 Hz$

Answer 2

The velocity of the wave is 364 m/s and the tension in the string is 0.578N

Calculating the tension in the string:

The velocity of a wave in a string is given by:

$v = \sqrt{\frac{T}{\mu}}$

where T is the tension in the string and μ is the mass per unit length

It is given that the string oscillates in its fundamental frequency, which is given by:

f = v/2L

239 = v/2×(0.762)

v = 364 m/s

Now, according to the relation between velocity of wave and the tension in the string, we get:

$v=\sqrt{\frac{T}{\mu} }\\\\T = v\mu^2$

T = 364×(0.762/0.48×10⁻³)

T = 0.578N

Learn more about fundamental frequency:

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