A string of length and mass is under tension . When the string vibrates, two successive harmonics are found to occur at frequencies and . The value of tension is ________ Newton. [2023]

Question

A string of length $1 m$ and mass $2 \times 10^{- 5} kg$ is under tension $T$ . When the string vibrates, two successive harmonics are found to occur at frequencies $750 Hz$ and $1000 Hz$ . The value of tension $T$ is ________ Newton. [2023]

Smart Achievers · Accepted Answer

(5)

As two successive harmonics are found to occur at frequency 750 Hz and 1000 Hz

$2 \times 10^{- 5} kg$

So, $750 = \frac{P}{2} \sqrt{\frac{T}{μ}} . . . (i)$

and, $1000 = \frac{P + 1}{2} \sqrt{\frac{T}{μ}} . . . (i i)$

Dividing eq. (ii) by (i),

$\frac{4}{3} = \frac{P + 1}{P}$

$∴$ $P = 3$

Putting this value of $P = 3$ in eq. (ii) and solving we get tension $T$ ,

$1000 = \frac{4}{2} \sqrt{\frac{T}{2 \times 10^{- 5}}}$

$∴$ $T = 5 N$

Solution

Q.
A string of length $1 m$ and mass $2 \times 10^{- 5} kg$ is under tension $T$ . When the string vibrates, two successive harmonics are found to occur at frequencies $750 Hz$ and $1000 Hz$ . The value of tension $T$ is ________ Newton. [2023]

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Solution

Q. A string of length 1 m and mass 2×10-5 kg is under tension T. When the string vibrates, two successive harmonics are found to occur at frequencies 750 Hz and 1000 Hz. The value of tension T is ________ Newton. [2023]

Ans. (5) As two successive harmonics are found to occur at frequency 750 Hz and 1000 Hz f=P2ℓTμ So, 750=P2Tμ ...(i) and, 1000=P+12Tμ ...(ii) Dividing eq. (ii) by (i), 43=P+1P ∴ P=3 Putting this value of P=3 in eq. (ii) and solving we get tension T, 1000=42T2×10-5 ∴ T=5 N

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Q.
A string of length $1 m$ and mass $2 \times 10^{- 5} kg$ is under tension $T$ . When the string vibrates, two successive harmonics are found to occur at frequencies $750 Hz$ and $1000 Hz$ . The value of tension $T$ is ________ Newton. [2023]