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enot [183]
3 years ago
5

Suppose you pluck a string on a guitar and it produces the note A at a frequency of 440 Hz. Now you press your finger down on th

e string against one of the frets, making this point the new end of the string. The newly shortened string has 4/5 the length of the full string. When you pluck the string, its frequency will be
A. 350 Hz
B. 440 Hz
C. 490 Hz
D. 550 Hz
Physics
1 answer:
MAVERICK [17]3 years ago
8 0

Answer:

D. 550 Hz

Explanation:

Frequency for string is given by

f_{} =\frac{v}{2L}

where v is the speed of the wave, L = m is the length

Frequency is inversely proportional to the Length of the string

f \alpha \frac{1}{L} \\

from above relation we can write

\frac{f_{1}}{f_{2}} =\frac{L_{2}}{L_{1}}

The newly shortened string has 4/5 the length of the full string .i,e

\frac{440}{f_{2}} =\frac{\frac{4}{5}L_{1} }{L_{1}}\\f_{2}=\frac{440\times5}{4} \\f_{2}=550 Hz

Hence option D is correct

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Hint: Here, we can see that 12 and 13 are consecutive numbers. So, all numbers between squares of 12 and 13 are non-square numbers. Therefore, first find squares of 12 and 13 and then subtract square of 12 from square of 13, we get numbers of non-square numbers. At the last subtract 1 from the result obtained as both extremes numbers are not included.

Complete step-by-step answer:

In these types of questions, a simple concept of numbers should be known that is between squares of two consecutive numbers all numbers are non-square numbers. Also one tricky point should remember that whenever we find the difference between two numbers we get a number of numbers between them including anyone of the extreme numbers. So we subtract 1 to exclude both extreme numbers.

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