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A String, Fixed at Both Ends, Vibrates in a Resonant Mode with a Separation of 2⋅0 Cm Between the Consecutive Nodes. for the Next Higher Resonant Frequency, this

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प्रश्न

A string, fixed at both ends, vibrates in a resonant mode with a separation of 2⋅0 cm between the consecutive nodes. For the next higher resonant frequency, this separation is reduced to 1⋅6 cm. Find the length of the string.

योग
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उत्तर

Given:
Separation between two consecutive nodes when the string vibrates in resonant mode = 2.0 cm
Let there be 'n' loops and
\[\lambda\] be the wavelength.

\[\lambda\] = \[2 \times Separation  between  the  consecutive  nodes\]

\[\lambda_1  = 2 \times 2 = 4  \text{ cm }\]

\[\lambda_2  = 2 \times 1 . 6 = 3 . 2  cm\]
Length of the wire is L.

In the first case:

\[L = \left( \frac{n \lambda_1}{2} \right)\] 

In the second case:

\[L = \left( n + 1 \right)\frac{\lambda_2}{2}\] 

\[ \Rightarrow \frac{n \lambda_1}{2} = \left( n + 1 \right)  \frac{\lambda_2}{2}\] 

\[ \Rightarrow n \times 4 = \left( n + 1 \right)\left( 3 . 2 \right)\] 

\[ \Rightarrow 4n - 3 . 2n = 3 . 2\] 

\[ \Rightarrow 0 . 8  n = 3 . 2\] 

\[ \Rightarrow n = 4
\text{ ∴ length of the string,}\]
\[L = \frac{\left( n \lambda_1 \right)}{2} = \frac{\left( 4 \times 4 \right)}{2} = 8  \text{ cm }\]

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अध्याय 15: Wave Motion and Waves on a String - Exercise [पृष्ठ ३२६]

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एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
अध्याय 15 Wave Motion and Waves on a String
Exercise | Q 42 | पृष्ठ ३२६

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