A string with a mass per length of 2.00 gm/m is stretched with a force of 150 N between two points that are 0.500 m apart.The fundamental frequency of the stretched string is in tune with the fundamental frequency of an organ pipe filled with air and open at both ends.The velocity of sound in air at 0$^{\circ}$C is 331 m/s.If the temperature of the air increases by 20$^{\circ}$C,then what is the length of the organ pipe?
A) 70.5 cm
B) 62.6 cm
C) 59.3 cm
D) 53.8 cm
E) 50.2 cm

Question

Quizplus · Accepted Answer

We can use the formula for the fundamental frequency of a stretched string:
$f=\frac{1}{2 L} \sqrt{\frac{T}{\mu}}$
where $L$ is the length of the string, $T$ is the tension, and $\mu$ is the mass per length.

Plugging in the given values, we have:
$f=\frac{1}{2 (0.500 	ext{ m})} \sqrt{\frac{(150 	ext{ N})}{(2.00 	ext{ gm/m})}}$
$f=\frac{1}{1.00 	ext{ m}} \sqrt{\frac{(150 	ext{ N})}{(0.00200 	ext{ kg/m})}}$
$f=\frac{1}{1.00 	ext{ m}} \sqrt{75000 	ext{ N/kg}}$
$f=273.9 	ext{ Hz}$

Since the fundamental frequency of the pipe is the same as the frequency of the string, we have:
$f=\frac{v}{2L}$
where $v$ is the velocity of sound in air.

Solving for $L$ and plugging in the given values, we have:
$L=\frac{v}{2f}=\frac{(331 	ext{ m/s} + 0.6 	imes 20 	ext{ m/s})}{2(273.9 	ext{ Hz})}=0.626 	ext{ m}=62.6 	ext{ cm}$

Therefore, the answer is (B) 62.6 cm.

A String with a Mass Per Length of 2 ∘^{\circ}∘ C Is 331 M/s

A String with a Mass Per Length of 2 $^{\circ}$ C Is 331 M/s