The frequency of a stretched uniform wire of length $L$ under tension is in resonance with the fundamental frequency of a closed pipe of same length. If the tension in the wire is increased by $8 \ N$,it is in resonance with the first overtone of the same closed pipe. The initial tension in the wire is (in $N$)

  • A
    $4$
  • B
    $0.5$
  • C
    $2$
  • D
    $1$

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Two tuning forks $A$ and $B$ give $5$ beats per second. Fork $A$ resonates with a closed air column $20 \ cm$ long and fork $B$ with an open air column $40.5 \ cm$ long. Neglecting end correction,the frequencies of tuning forks $A$ and $B$ are respectively.

Match Column-$1$ with Column-$2$.
Column-$1$Column-$2$
$(a)$ Light waves$(i)$ Mechanical and transverse.
$(b)$ Sound waves$(ii)$ Mechanical and longitudinal.
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$(d)$ Waves on a tense string$(iv)$ Mechanical,transverse and longitudinal.

$A$ narrow tube is bent in the form of a circle of radius $R,$ as shown in the figure. Two small holes $S$ and $D$ are made in the tube at positions right-angled to each other. $A$ source placed at $S$ generates a wave of intensity $I_0$ which is equally divided into two parts: one part travels along the longer path,while the other travels along the shorter path. Both the waves meet at the point $D$ where a detector is placed. The maximum intensity produced at $D$ is given by

$A$ 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:

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Select the correct alternative$(s)$ :-
$(A)$ Number of nodes equals to number of antinodes in closed organ pipe.
$(B)$ In open organ pipe,if number of antinodes is $m$,then number of nodes will be $m-1$.
$(C)$ If frequency of $4^{\text{th}}$ harmonic of open organ pipe is $400 \ Hz$,then frequency of $2^{\text{nd}}$ overtone of closed organ pipe of same length is $250 \ Hz$.
$(D)$ Time interval between successive maxima or minima (for superposition of two waves) is $\Delta t = \frac{1}{|f_1-f_2|} \ s$.

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