A stretched wire carries a body of density??=8000kg/m3?at its end. The fundamental frequency?of vibration of wire is 280 Hz. The body is dipped?completely in a vessel of water. Find the new?frequency of fundamental mode of vibrations.?(Density of water isr=1000kg/m3)

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Solution

$F u n d a m e n t t a l f r e q u e n c y i s g i v e n b y f_{0} = \frac{1}{2 l} \sqrt{\frac{T}{μ}} I n f i r s t c a s e t e n s i o n i s b a l a n c e d b y w e i g h t, W \Rightarrow T = W = ρ V g T = 8000 V g f_{0} = \frac{1}{2 l} \sqrt{\frac{8000 V g}{μ}} G i v e n : f_{0} = 280 H z 280 = \frac{1}{2 l} \sqrt{\frac{8000 V g}{μ}}____________(1) W h e n d i p p e d i n w a t e r a n a d d i t i o n a l b u o y a n t f o r c e (F_{B} = ρ_{w a t e r} V g) a c t s u p w a r d T' + F_{B} = W T' = W - F_{B} T' = 8000 V g - 1000 V g T' = 7000 V g N e w f u n d a m e n t a l f r e q u e n c y f_{o}' = \frac{1}{2 l} \sqrt{\frac{T'}{μ}} f_{o}' = \frac{1}{2 l} \sqrt{\frac{7000 V g}{μ}}__________(2) D i v i d i n g e q u a t i o n 1 a n d 2 \frac{f_{o}'}{280} = \frac{\frac{1}{2 l} \sqrt{\frac{7000 V g}{μ}}}{\frac{1}{2 l} \sqrt{\frac{8000 V g}{μ}}} = \frac{\sqrt{\frac{7000 V g}{μ}}}{\sqrt{\frac{8000 V g}{μ}}} = \frac{\sqrt{7}}{\sqrt{8}} \frac{f_{o}'}{280} = \sqrt{\frac{7}{8}} f_{o}' = 280 \sqrt{\frac{7}{8}} = 261.92 H z$

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