Question

A sonometer wire stretched by a load of density p has a fundamental frequency n1. When th load is completely immersed in water its fundamental frequency is n2. Show that n2/n1 =√p-pw /p

Answer 1

Answer:

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Answer 2

Hence proved that n2 / n1 = √ρ − ρ (w) / ρ

Explanation:

The relative density is given as:

• Relative density "s" = ρ / ρ(w)
• Relative density "s"  = W1 /  W 1 − W 2
• Relative density "s" =  T1 /T 1 − T 2

Now

T1 / T2 = ρ / ρ − ρ (w)

n1 = 1 / 2 L √T1 / m and

n2 = 1 / 2 L √  T2 / m

n2 / n1 = T2 / T1

n2 / n1 = ρ − ρ (w) / ρ

Hence proved that n2 / n1 = √ρ − ρ (w) / ρ