A tuning fork $A$, marked 512 Hz , produces 5 beats per second, where sounded with another unmarked tuning fork $B$. If $B$ is loaded with wax the number of beats is again 5 per second. What is the frequency of the tuning fork $B$ when not loaded?
Frequency of tuning fork $A$,
$$v_A=512 \mathrm{~Hz}$$
Probable frequency of tuning fork $B$,
$$v_B=v_A \pm 5=512 \pm 5=517 \text { or } 507 \mathrm{~Hz}$$
when $B$ is loaded, its frequency reduces.
If it is 517 Hz , it might reduced to 507 Hz given again a beat of 5 Hz .
If it is 507 Hz , reduction will always increase the beat frequency, hence $v_B=517 \mathrm{~Hz}$
The displacement of an elastic wave is given by the function $y=3 \sin \omega t+4 \cos \omega t$, where $y$ is in cm and $t$ is in second. Calculate the resultant amplitude.
Given, displacement of an elastic wave $y=3 \sin \omega t+4 \cos \omega t$
Assume,
$$\begin{aligned} & 3=a \cos \phi \quad \text{... (i)}\\ & 4=a \sin \phi \quad \text{... (ii)} \end{aligned}$$
$$\begin{aligned} &\text { On dividing Eq. (ii) by Eq. (i) }\\ &\tan \phi=\frac{4}{3} \Rightarrow \phi=\tan ^{-1}(4 / 3) \end{aligned}$$
$$\begin{array}{ll} \text { Also, } & a^2 \cos ^2 \phi+a^2 \sin ^2 \phi=3^2+4^2 \\ \Rightarrow & a^2\left(\cos ^2 \phi+\sin ^2 \phi\right)=25 \end{array}$$
$$a^2 \cdot 1=25 \Rightarrow a=5$$
$$\begin{aligned} \text{Hence,}\quad Y & =5 \cos \phi \sin \omega t+5 \sin \phi \cos \omega t \\ & =5[\cos \phi \sin \omega t+\sin \phi \cos \omega t]=5 \sin (\omega t+\phi) \end{aligned}$$
where $\quad \phi=\tan ^{-1}(4 / 3)$
Hence, amplitude $=5 \mathrm{~cm}$
A sitar wire is replaced by another wire of same length and material but of three times the earlier radius. If the tension in the wire remains the same, by what factor will the frequency change?
$$\begin{aligned} &\text { Frequency of vibrations produced by a stretched wire }\\ &\mathrm{v}=\frac{n}{2 l} \sqrt{\frac{T}{\mu}} \end{aligned}$$
$$\text { Mass per unit length } \mu=\frac{\text { Mass }}{\text { Length }}=\frac{\pi r^2 l \rho}{l}=\pi r^2 \rho \quad\left[\because M=v p=A l \rho=\pi r^2 l \rho\right]$$
$$\begin{aligned} &\begin{aligned} \therefore \quad v & =\frac{n}{2 l} \sqrt{\frac{T}{\pi r^2 \rho}} \Rightarrow \mathrm{v} \propto \sqrt{\frac{1}{r^2}} \\ & \mathrm{v} \propto \frac{1}{r} \end{aligned}\\ &\text { Hence, when radius is tripled, } v \text { will be } \frac{1}{3} \text { rd of previous value. } \end{aligned}$$
At what temperatures (in ${ }^{\circ} \mathrm{C}$ ) will the speed of sound in air be 3 times its value at $0^{\circ} \mathrm{C}$ ?
We know that speed of sound in air $v \propto \sqrt{T}$
$\therefore \frac{v_T}{v_0}=\sqrt{\frac{T_T}{T_0}}=\sqrt{\frac{T_T}{273}}\quad$ [where $T$ is in kelvin]
$$\begin{aligned} \text{But}\quad & \frac{v_T}{v_0}=\frac{3}{1} \quad [\because \text{ speed becomes three times]}\\ \therefore\quad & \frac{3}{1}=\sqrt{\frac{T_T}{T_0}} \Rightarrow \frac{T_T}{273}=9 \\ \therefore \quad &T_T=273 \times 9=2457 \mathrm{~K} \\ & =2457-273=2184^{\circ} \mathrm{C} \end{aligned}$$
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
$$\begin{aligned} & \quad v_b=n_1-n_2 \\ & \therefore \quad \text { Time period of beats }=T_b=\frac{1}{v_b}=\frac{1}{n_1-n_2} \end{aligned}$$