HSC Science (Electronics) 12th Board ExamMaharashtra State Board
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# Solution for A Stretched Wire Emits a Fundamental Note of Frequency 256 Hz. Keeping the Stretching Force Constant and Reducing the Length of Wire by 10 Cm, the Frequency Becomes 320 Hz. Calculate the Original Length of Wire. - HSC Science (Electronics) 12th Board Exam - Physics

#### Question

A stretched wire emits a fundamental note of frequency 256 Hz. Keeping the stretching force constant and reducing the length of wire by 10 cm, the frequency becomes 320 Hz. Calculate the original length of wire.

#### Solution

Let ‘l’ be the length of the wire which emits a fundamental note of frequency 256 Hz. When length = (l − 10) cm, fundamental frequency n = 320 Hz.
We know that the fundamental frequency n of a stretched string is given by

n=1/(2l)sqrt("T"/"m")

where ‘T’ is the tension and ‘m’ the linear density of the string.

When length = l, n = 256 Hz

i.e.

256=1/(2l)sqrt("T"/"m")......................(1)

When length = (l − 10) cm, n = 320 Hz

320=1/(2(l-10))sqrt("T"/"m")..........................................(2)

Dividing (1) by (2) gives

256/320=(2(l-10))/(2l)

4/5=[(l-10)]/l

l 50 cm = 0.5 m

Therefore, the original length of the wire is 50 cm.

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Solution for question: A Stretched Wire Emits a Fundamental Note of Frequency 256 Hz. Keeping the Stretching Force Constant and Reducing the Length of Wire by 10 Cm, the Frequency Becomes 320 Hz. Calculate the Original Length of Wire. concept: Free and Forced Vibrations. For the courses HSC Science (Electronics), HSC Science (Computer Science), HSC Science (General)
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