Question

if the length of a stretched string is shortened by 20% and the tension is increased by 21% then the frequency is??​

Answer 1

Explanation:

Given If the length of a stretched string is shortened by 20% and the tension is increased by 21% then the frequency is?

We know that fundamental frequency n = 1/2l √T/m

So let initial frequency be n1 and final frequency be n2

Now n1 = 1/2l 1 √T1 / m

Also n2 = 1/2l 2 √T2 /m

Dividing both the equation we get  

So n1 / n2 = l2 / l1 √T1/T2

Now n2/n1 = l 1 / l2 √T2 / T1

                  = l 1 / [l1 – 20/100] √T1 + 21 / 100 T1 / T1

                = 100 l1 / l1 x 11 / 10

So frequency n2 : n1 = 11:8

Answer 2

Answer:37.5

Explanation:check attachment