General Physics Problem - Frequency of First Overtone?

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m w

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the fundamental vibrational mode of the string would be...
f = v / 2L

the first overtone would be
f = v / L

where
f = frequency
v = wave velocity
L = length of the string.

and ...
v = ( T / (m / L))^.5

where
T = tension
m = mass
L = length of the string.

see the reference for details...

************
so let's start by finding m... mass of the string...

we're only interested in the part of the string that is vibrating. from the post to the bridge. so why were we given total length of the string and total mass? the answer is we use that to calculate the mass in the part that is vibrating...

92 cm = 3.2 g.. right?.. therefore...
70 cm x (3.2 g / 92 cm) = 2.43 g.. that is the mass of the "vibrating" part of the string

then T and L in the correct units...
T = 480 N = 480 kg m / s² = 4.8x10^5 g m / s²
L = 70 cm = 70 cm x ( 1 m / 100 cm) = 0.70 m

so that...
v = ((4.8x10^5 g m / s²) / (2.43 g / 0.70 m))^.5 = (1.38x10^5 m²/s²)^.5
v = 371.8 m/s

*****
next...

f = v / L = (371.8 m/s) / (0.70 m) = 531 /s = 531 Hz

************
************
my guess is that you did this...

v = (4.8x10^5 / (3.2 / 0.92 ))^.5
v = 371.5 m/s
and
f = v / 2L = 371.5 / (2 x 0.70 m) = 265 Hz...

right? or you may have calculated v correctly.. either way... it looks like your mistake was the fundamental vs first overtone.. v / 2L vs v / L...

***********
questions?
 
the fundamental vibrational mode of the string would be...
f = v / 2L

the first overtone would be
f = v / L

where
f = frequency
v = wave velocity
L = length of the string.

and ...
v = ( T / (m / L))^.5

where
T = tension
m = mass
L = length of the string.

see the reference for details...

************
so let's start by finding m... mass of the string...

we're only interested in the part of the string that is vibrating. from the post to the bridge. so why were we given total length of the string and total mass? the answer is we use that to calculate the mass in the part that is vibrating...

92 cm = 3.2 g.. right?.. therefore...
70 cm x (3.2 g / 92 cm) = 2.43 g.. that is the mass of the "vibrating" part of the string

then T and L in the correct units...
T = 480 N = 480 kg m / s² = 4.8x10^5 g m / s²
L = 70 cm = 70 cm x ( 1 m / 100 cm) = 0.70 m

so that...
v = ((4.8x10^5 g m / s²) / (2.43 g / 0.70 m))^.5 = (1.38x10^5 m²/s²)^.5
v = 371.8 m/s

*****
next...

f = v / L = (371.8 m/s) / (0.70 m) = 531 /s = 531 Hz

************
************
my guess is that you did this...

v = (4.8x10^5 / (3.2 / 0.92 ))^.5
v = 371.5 m/s
and
f = v / 2L = 371.5 / (2 x 0.70 m) = 265 Hz...

right? or you may have calculated v correctly.. either way... it looks like your mistake was the fundamental vs first overtone.. v / 2L vs v / L...

***********
questions?
 
A guitar string is 92 cm long and has a mass of 3.2 g. From the bridge to the support post (L) is 70 cm, and the string is under a tension of 480 N. What is the frequency of the first overtone (in Hertz)?

Please help!!!
My first answer, 265.35 Hz, was not right. Looking for final numerical answer and a short explanation. Thank you so much!
 
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