Physics, 28.07.2019 18:00, londonpaige

Two strings on a musical instrument are tuned to play at 196 hz (g) and 523 hz (c). (a) what are the first two overtones for each string? first overtone for 196 hz (g) hz second overtone for 196 hz (g) hz first overtone for 523 hz (c) hz second overtone for 523 hz (c) hz (b) if the two strings have the same length and are under the same tension, what must be the ratio of their masses (m196/m523)? (c) if the strings, instead, have the same mass per unit length and are under the same tension, what is the ratio of their lengths (l196/l523)? 1.33 2.67 (d) if their masses and lengths are the same, what must be the ratio of the tensions in the two strings? (t196/t523) 45 0.14

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(a) first two overtones for each string:
The first string has a fundamental frequency of 196 Hz. The n-th overtone corresponds to the (n+1)-th harmonic, which can be found by using
f_n = n f_1

where f1 is the fundamental frequency.

So, the first overtone (2nd harmonic) of the string is
$f_2 = 2 f_1 = 2 \cdot 196 Hz = 392 Hz$
while the second overtone (3rd harmonic) is
$f_3 = 3 f_1 = 3 \cdot 196 Hz = 588 Hz$

Similarly, for the second string with fundamental frequency f_1 = 523 Hz

, the first overtone is
$f_2 = 2 f_1 = 2 \cdot 523 Hz = 1046 Hz$
and the second overtone is
$f_3 = 3 f_1 = 3 \cdot 523 Hz = 1569 Hz$

(b) The fundamental frequency of a string is given by
$f= \frac{1}{2L} \sqrt{ \frac{T}{\mu} }$
where L is the string length, T the tension, and $\mu = m/L$ is the mass per unit of length. This part of the problem says that the tension T and the length L of the string are the same, while the masses are different (let's calle them $m_{196}$ , the mass of the string of frequency 196 Hz, and $m_{523}$ , the mass of the string of frequency 523 Hz.
The ratio between the fundamental frequencies of the two strings is therefore:
$\frac{523 Hz}{196 Hz} = \frac{ \frac{1}{2L} \sqrt{ \frac{T}{m_{523}/L} } }{\frac{1}{2L} \sqrt{ \frac{T}{m_{196}/L} }}$
and since L and T simplify in the equation, we can find the ratio between the two masses:
$\frac{m_{196}}{m_{523}}=( \frac{523 Hz}{196 Hz} )^2 = 7.1$

(c) Now the tension T and the mass per unit of length $\mu$ is the same for the strings, while the lengths are different (let's call them $L_{196}$ and $L_{523}$ ). Let's write again the ratio between the two fundamental frequencies
$\frac{523 Hz}{196 Hz}= \frac{ \frac{1}{2L_{523}} \sqrt{ \frac{T}{\mu} } }{\frac{1}{2L_{196}} \sqrt{ \frac{T}{\mu} }}$
And since T and $\mu$ simplify, we get the ratio between the two lengths:
$\frac{L_{196}}{L_{523}}= \frac{523 Hz}{196 Hz}=2.67$

(d) Now the masses m and the lenghts L are the same, while the tensions are different (let's call them $T_{196}$ and $T_{523}$ . Let's write again the ratio of the frequencies:
$\frac{523 Hz}{196 Hz}= \frac{ \frac{1}{2L} \sqrt{ \frac{T_{523}}{m/L} } }{\frac{1}{2L} \sqrt{ \frac{T_{196}}{m/L} }}$
Now m and L simplify, and we get the ratio between the two tensions:
$\frac{T_{196}}{T_{523}}=( \frac{196 Hz}{523 Hz} )^2=0.14$

answered: Guest

Hypothesis. why don’t you know this? i learned this stuff in like 5th grade

answered: Guest

it is given that trillions of microbes make up to 1% to 3% of the body mass.

let's take average of this percentage: 2%

consider the average mass of the human body as 70 kg.

then, trillions of microbes would have mass $= 2\%\hspace{1mm}of \hspace{1mm} 70\hspace{1mm} kg=\frac {2}{100}\times 70 =1.4\hspace{1mm} kg$