Question

when the string is increased by 44%.the frequency increased by 10hz the frequency of the string is

Answer 1

This is probably related to simple harmonic motion and standing waves on a wire under load/tension T. The frequency f is the fundamental oscillating frequency.

Frequency f = k × sqrt(T)

Differentiate:

so delta f = k × 1/[2 sqrt(T)] × delta T

delta f / f = 1/[2 T] × delta T

= 1/2 × 44% = 0.22

Substituting given data values we get

f = delta f ÷ 0.22

= 10 Hz / 0.22

Frequency f = 45.55 Hz

Answer 2

Explanation:

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