A string is fixed at both ends vibrates in a resonant mode with a separation 2.0 ,,cm between consec

English

Gujarati

Hindi

14.Waves and Sound

hard

A string is fixed at both ends vibrates in a resonant mode with a separation $2.0 \,\,cm$ between the consecutive nodes. For the next higher resonant frequency, this separation is reduced to $1.6\,\, cm$. The length of the string is .... $cm$

A

$4$

B

$8$

C

$12$

D

$16$

Solution

Let there be $n$ loops in the $1\, st$ case

$\rightarrow$ Length of the wire$L$ $=\left(\left(\mathrm{n} \lambda_{1}\right) / 2\right)$

$\left(\lambda_{1}=2 \times 2=4 \mathrm{cm}\right)$

$\rightarrow$ Length of the wire $L_{l}=\left\{(n+1) \frac{\lambda_{2}}{2}\right\}$

$\left(\lambda_{2}=2 \times(1.6)=3.2 \mathrm{cm}\right)$

$\rightarrow \frac{n \lambda_{1}}{2}=(n+1) \frac{\lambda_{2}}{2}$

$\rightarrow n \times 4=(n+1)(3.2)$

$\rightarrow 4 n-(3.2) n=3.2$

$\rightarrow 0.8 n=3.2$

$\rightarrow n=4$

Length of the string

$L=\frac{n \lambda_{1}}{2}=\frac{(4 \times 4)}{2}=8 \mathrm{cm}$

Standard 11

Physics

Share

Similar Questions

A massless rod of length $L$ is suspended by two identical strings $AB$ and $CD$ of equal length. A block of mass $m$ is suspended from point $O$ such that $BO$ is equal to $‘x’$. Further it is observed that the frequency of $1^{st}$ harmonic in $AB$ is equal to $2^{nd}$ harmonic frequency in $CD$. $‘x’$ is

hard

A vibrating string of certain length $l$ under a tension $T$ resonates with a mode corresponding to the first overtone (third harmonic) of an air column of length $75\, cm$ inside a tube closed at one end. The string also generates $4\, beats$ per second when excited along with a tuning fork of frequency $n$. Now when the tension of the string is slightly increased the number of beats reduces to $2\, per second$. Assuming the velocity of sound in air to be $340\, m/s$, the frequency $n$ of the tuning fork in $Hz$ is

hard

If the tension of sonometer’s wire increases four times then the fundamental frequency of the wire will increase by …. $ times$

easy

A guitar string of length $90\,cm$ vibrates with a fundamental frequency of $120\,Hz.$ The length of the string producing a fundamental frequency of $180\,Hz$ will be $………..cm$.

easy

(JEE MAIN-2023)

Explain the reflection of wave at free support.

medium