Question

a stretched wire emits a fundamental note of 384Hz if the length of wire is reduced of 10cm keeping the tension constant, the frequency becomes 512Hz calculate the original length of the wire.​

Answer 1

di ko to be a good time⏰

Answer 2

Explanation:

Given A stretched wire emits a fundamental note of 384 Hz if the length of wire is reduced of 10 cm keeping the tension constant, the frequency becomes 512 Hz calculate the original length of the wire.

• Let l be the length of the wire which emits a fundamental note of frequency 384 hz. So length = (l – 10) cm, fundamental frequency n = 512 hz
• We know that fundamental frequency n of a stretched string is
• n = 1/2l √T/m where T is tension and m is linear density of the string.
• 384 = 1/2l √T/m ------------1
• 512 = 1/2(l – 10)√T/m ----------2
• Dividing 1 by 2 we get
• 384 / 512 = 2 (l – 10) / 2l
• 3/4 = l – 10 / l
• 3l = 4l – 40
• Or l = 40 cm = 0.4 m

Therefore original length of the wire is 40 cm or 0.4 m

Reference link will be

https://brainly.in/question/2704246