A sonometer wire of resonating length 90 ~cm has a fundamental frequency of 400 ~Hz when kept under some tension. The resonating length of the wire with fundamental frequency of 600 ~Hz under same tension _____ cm .

Correct answer: 0 aligned & f_0=400 Hz ; v= T μ = constant & (λ)/(2)=L ; v=f_0 λ aligned aligned & v 2 f_0 =L v=2 Lf_0 & L^ = v 2 f^ = 2 Lf_0 2 f^ & = Lf_0 f^ =(90 × 400)/(600)=60 aligned

JEE Main 20245 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 5 April
Shift: Evening Shift
Subject: Physics
Chapter: Waves and Sound

JEE Main 2024 · 5 April · Evening Shift · Physics · Waves and Sound

A sonometer wire of resonating length 90 ~cm has a fundamental...

A sonometer wire of resonating length

90 \mathrm{~cm}

has a fundamental frequency of

400 \mathrm{~Hz}

when kept under some tension. The resonating length of the wire with fundamental frequency of

600 \mathrm{~Hz}

under same tension _____

\mathrm{cm}

Correct answer

Step-by-step solution

\begin{aligned} & \mathrm{f}_0=400 \mathrm{ Hz} ; \mathrm{v}=\sqrt{\frac{\mathrm{T}}{\mu}}=\text{ constant } \\ & \frac{\lambda}{2}=\mathrm{L} ; \quad \mathrm{v}=\mathrm{f}_0 \lambda\end{aligned}

\begin{aligned} & \frac{\mathrm{v}}{2 \mathrm{f}_0}=\mathrm{L} \Rightarrow \mathrm{v}=2 \mathrm{Lf}_0 \\ & \mathrm{ L}^{\prime}=\frac{\mathrm{v}}{2 \mathrm{f}^{\prime}}=\frac{2 \mathrm{Lf}_0}{2 \mathrm{f}^{\prime}} \\ & =\frac{\mathrm{Lf}_0}{\mathrm{f}^{\prime}}=\frac{90 \times 400}{600}=60\end{aligned}

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