When two waves of almost equal frequencies $n_1$ and $n_2$ reach at a point simultaneously, what is the time interval between successive maxima?
Let, $$n_1>n_2$$
Beat frequency
$$\begin{aligned} & \quad v_b=n_1-n_2 \\ & \therefore \quad \text { Time period of beats }=T_b=\frac{1}{v_b}=\frac{1}{n_1-n_2} \end{aligned}$$
A steel wire has a length of 12 m and a mass of 2.10 kg . What will be the speed of a transverse wave on this wire when a tension of $2.06 \times 10^4 \mathrm{~N}$ is applied?
$$\begin{aligned} &\text { Given, length of the wire }\\ &l=12 \mathrm{~m} \end{aligned}$$
Mass of wire $$m=2.10 \mathrm{~kg}$$
Tension $T=2.06 \times 10^4 \mathrm{~N}$
Speed of transverse wave $\quad v=\sqrt{\frac{T}{\mu}}$ [where $\mu=$ mass per unit length]
$$=\sqrt{\frac{2.06 \times 10^4}{\left(\frac{2.10}{12}\right)}}=\sqrt{\frac{2.06 \times 12 \times 10^4}{2.10}}=343 \mathrm{~m} / \mathrm{s}$$
A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a source of 1237.5 Hz ? (sound velocity in air $=330 \mathrm{~ms}^{-1})$
Length of pipe
$$\begin{aligned} & v_{\text {funda }}=\frac{V}{4 L}=\frac{330}{4 \times 20 \times 10^{-2}} \quad \text{(for closed pipe)}\\ & v_{\text {funda }}=\frac{330 \times 100}{80}=412.5 \mathrm{~Hz} \\ & \frac{v_{\text {given }}}{v_{\text {funda }}}=\frac{1237.5}{412.5}=3 \end{aligned}$$
Hence, 3rd harmonic node of the pipe is resonantly excited by the source of given frequency.
A train standing at the outer signal of a railway station blows a whistle of frequency 400 Hz still air. The train begins to move with a speed of $10 \mathrm{~ms}^{-1}$ towards the platform. What is the frequency of the sound for an observer standing on the platform? (sound velocity in air $=330 \mathrm{~ms}^{-1}$ )
As the source (train) is moving towards the observer (platform) hence apparent frequency observed is more than the natural frequency.
Frequency of whistle $\nu=400 \mathrm{~Hz}$
Speed of train $v_t=10 \mathrm{~m} / \mathrm{s}$
Velocity of sound in air $v=330 \mathrm{~m} / \mathrm{s}$
$$\text { Apparent frequency when source is moving } v_{\mathrm{app}}=\left(\frac{v}{v-v_t}\right) v$$
$$\begin{aligned} = & \left(\frac{330}{330-10}\right) 400 \\ \Rightarrow \quad v_{\text {app }} & =\frac{330}{320} \times 400=412.5 \mathrm{~Hz} \end{aligned}$$
The wave pattern on a stretched string is shown in figure. Interpret what kind of wave this is and find its wavelength.
We have to observe the displacement and position of different points, then accordingly nature of two wave is decided.
Points on positions $x=10,20,30,40$ never move, always at mean position with respect to time. These are forming nodes which characterise a stationary wave.
$$\begin{aligned} & \because \text { Distance between two successive nodes }=\frac{\lambda}{2} \\ & \begin{aligned} \Rightarrow \quad \lambda & =2 \times(\text { node to node distance }) \\ & =2 \times(20-10) \\ & =2 \times 10=20 \mathrm{~cm} \end{aligned} \end{aligned}$$