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Question #105446
At a temperature of 27∘C. two identical organ pipes produce notes of frequency 140 Hz. If the temperature of one pipe is raised to 57.75∘C. the number of beats produced per second is
Ans: 7 beats/s
Expert's answer
v
=
γ
R
T
M
v=\sqrt{\frac{\gamma RT}{M}}
v
=
M
γ
RT
v
2
v
1
=
T
2
T
1
\frac{v_2}{v_1}=\sqrt{\frac{T_2}{T_1}}
v
1
v
2
=
T
1
T
2
f
+
x
f
=
T
2
T
1
\frac{f+x}{f}=\sqrt{\frac{T_2}{T_1}}
f
f
+
x
=
T
1
T
2
140
+
x
140
=
273.15
+
57.75
273.15
+
27
\frac{140+x}{140}=\sqrt{\frac{273.15+57.75}{273.15+27}}
140
140
+
x
=
273.15
+
27
273.15
+
57.75
x
=
7
b
e
a
t
s
s
x=7\frac{beats}{s}
x
=
7
s
b
e
a
t
s
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