G L Mittal and TARUN MITTAL Solutions for Chapter: Vibrations of Stretched Strings, Exercise 3: FOR DIFFERENT COMPETITIVE EXAMINATIONS

A tuning fork of frequency $480 \mathrm{hertz}$ vibrating with a sonometer wire produces $10 \mathrm{beats} \mathrm{per} \sec .$ If on lightening the wire the number of beats per second decreases, then the frequency of the wire will be :

Select all the correct alternatives. Two identical straight wires are stretched so as to produce $6 \mathrm{beats} \mathrm{per} \mathrm{second}$ when vibrating simultaneously. On changing the tension slightly in one of them, the beat-frequency remains unchanged. Denoting by $T_1, T_2$ the higher and the lower initial tensions in the strings, then it could be said that while making the above changes in tension.

A horizontal stretched string fixed at two ends, is vibrating in its fifth harmonic according to the equation $\mathrm{y}\left(\mathrm{x},\mathrm{t}\right)=0.01 \mathrm{m} \sin \left[\left(62.8 {\mathrm{m}}^{-1}\right)\mathrm{x}\right] \cos \left[\left(628 {\mathrm{s}}^{-1}\right)\mathrm{t}\right].$ Assuming $\mathrm{\pi }=3.14,$ the correct statement is (are)

A sonometer wire, loaded by a mass of $9 \mathrm{kg},$ resonates with a given tuning fork and five antinodes are observed in the stationary waves formed between the two bridges. When a mass $\mathrm{M}$ replaces the $9 \mathrm{kg}$ mass, then with the same tuning fork and in the same positions of the bridges, three antinodes are observed. The mass $\mathrm{M} \mathrm{is} :$

A wire of length $\mathrm{l},$ radius $\mathrm{r}$ and density $\mathrm{d}$ has a tension $\mathrm{T}.$ The estimated graphs for the frequency $\mathrm{n}$ of the wire are drawn. Explain with reason which graph is correct :

A vibrating string of certain length $\mathrm{l}$ under a tension $\mathrm{T}$ resonates with a mode corresponding to the first overtone (third harmonic) of an air column of length $75 \mathrm{cm}$ inside a tube closed at one end. The string also generates $4 \mathrm{beats} \mathrm{per} \mathrm{second}$ when excited, along with a tuning fork of frequency $\mathrm{n}$. Now when the tension of the string is slightly increased, the number of beats reduces to $2 \mathrm{per} \mathrm{second}$.  Assuming the velocity of sound in air to be $340 \mathrm{m} {\mathrm{s}}^{-1}$ the frequency $\mathrm{n}$ of the tuning fork is $\mathrm{Hz} \mathrm{is} :$

The equation of a wave on a string of linear mass density $0.04 \mathrm{kg} {\mathrm{m}}^{-1}$ is given by $\mathrm{y}=0.22 \sin 2\mathrm{\pi }\left(\frac{\mathrm{t}}{0.04}-\frac{\mathrm{x}}{0.50}\right)$ where distances are measured in meters and time in seconds. The tension in the string is:

A hollow pipe of length $0.8 \mathrm{m}$ is closed at one end. At its open end a $0.5 \mathrm{m}$ long uniform string is vibrating in its second harmonic and its resonates with the fundamental frequency of the pipe, If the tension in the wire is $50 \mathrm{N}$ and the speed of sound is $320 {\mathrm{ms}}^{-1}$, the mass of the string is :