$\Delta p=A \operatorname{Sin} k x \operatorname{Cos} \omega t \Rightarrow \Delta p(x=0)=0$

$\Delta p(x=L)=0 \Rightarrow \operatorname{Sin} k L=0$

$k=\frac{n \pi}{L}$

$\Delta p(x, t)=A \operatorname{Sin} k x \operatorname{Cos} \omega t$

$\Delta p(x=L, t)=0 \quad \operatorname{SinkL}=0$

$k=\frac{n \pi}{L}$

$\frac{2 \pi}{\lambda}=\frac{n \pi}{L} \Rightarrow \lambda \cdot\left(\frac{2 L}{n}\right)$

$\nu_n=\frac{v}{\lambda}=n\left(\frac{v}{2 L}\right)$

$\nu_1=\frac{v}{2 L},$

$\nu_2 = 2 \cdot\left(\frac{v}{\mu}\right) \cdots$

$\nu = n\left(\frac{v}{2 L}\right)$ Harmonic

$n=1$ First Harmonic

$n=2$ second Harmonic

$\left(\frac{V}{2 L}\right)=$ fundamental frequency

$\Rightarrow \quad \nu_n=n\left(\frac{v}{2 L}\right)$

$v=\sqrt{\frac{T}{r}}$

$v=\gamma \sqrt{\frac{B}{\rho}}$

$\nu = n\left(\frac{v}{2L}\right)$ Harmonic

$n=1$ First Harmonic

$n=2$ second Harmonic

$\left(\frac{V}{2 L}\right)=$ Fundamental frequency

$\Rightarrow \quad \nu_n=n\left(\frac{v}{2 L}\right)$

$v=\sqrt{\frac{T}{r}}$

$v=\gamma \sqrt{\frac{B}{\rho}}$

$y(x)=A \operatorname{Sink} \operatorname{Cos} \omega t$

$\sin (k L) \neq 0$ it is maxismum possible displacement

$k L =(2 n+1) \frac{\pi}{2}$ = $\frac{2 \pi}{\lambda} \cdot L =(2 n+1) \frac{\pi}{2}$

$\lambda =\frac{4 L}{(2 n+1)}$

$\Delta p(x, t)=A \operatorname{Sin} k x \cos \omega t$

$\left.k x\right|_L=k L=(2 n+1) \frac{\pi}{2}$

$\frac{2 \pi}{\lambda} L=(2 n+1) \frac{\pi}{2} \rightarrow \lambda=\frac{4 L}{(2 n+1)}$

$\nu_n=\frac{v}{\lambda}=\frac{(2 n+1)}{4 L} v=(n+\frac{1}{2})(\frac{v}{2 L})v=\gamma\sqrt{\frac{B}{\rho}}$

B = Bulk modulus

$\rho$ = density

$\gamma=C_p / C_v$

$n \frac{\lambda}{2}=L$

$\Rightarrow \quad \lambda=(\frac{2 L}{n})$

One difference between the string vibration and air column vibration.

END CORRECTION

R = radius of the pipe

EFFECTIVE Length for ended pide = L+1.2R

EFFECTIVE length for a pipe closed at one and = L+0.6R

Superposition of two waves of tHE SAME FREQUENCY

traveling in opposite directions

Standing wave.

SUPERPOSITION OF WAVES THAT HAVE SLIGHTLY DIFFERENT FREQUENCIES

$\nu_1 \quad \nu_2$

$|(\nu_1-\nu_2)| \ll \nu_1 \sim \nu_2$

$\nu_1=500 {Hz},$

$\nu_2=502 {Hz}$

BEAT PHENOMENA

$y_1(x, t)=A \sin (k_1 x-\omega_1 t)$

$y_2(x, t)=B \sin (k_2 x-\omega_2 t)$

We stand at a point $x = x_0 = 0$

$y_1(t)=+A \sin \omega_1 t$

$y_2(t)=B \sin \omega_2 t$

$y(t)=y_1(t)+y_2(t)=A \sin \omega_1 t+B \sin \omega_2 t$

$y(t)= A \sin \omega_1 t+B \sin \omega_2 t$

Beats can be understand easily if we take A =B

$y(t)=A[\sin \omega_1 t+\sin \omega_2 t]=2 A \sin(\frac{\omega_1+\omega_2}{2} t) \cos(\frac{\omega_1-\omega_2}{2} t)$

$If \omega_1=\omega_2 \quad y(t) \approx 2 A \sin \omega t$

$\omega_1 \neq \omega_2 \quad \omega_2=\omega_1+\Delta \omega \quad \Delta \omega \ll \omega_1$

$y(t)=2 A \sin \omega_{+} t \cdot \cos \omega_{-} t$

$\omega_{+}=\frac{\omega_1+\omega_2}{2} \simeq \omega_1,$

$\omega_{-}=|(\omega_1-\omega_2)|$

$y(t)=\quad 2 A \sin \omega_{+} t \cos \omega_{-} t$

$\omega_{-} \ll \omega_{+}$

$T_{+}=\frac{2 \pi}{\omega_{+}} \ll T_{-}=\frac{2 \pi}{\omega_{-}}$

$y(t)=2 A \sin \omega_{+} t c_0 \omega_{-} t$

Phenomena of Beats

$\Delta p(x, t)=P \sin \omega_{+t}{\cos \omega_{-t}}$

• ${(\frac{\omega_1-\omega_2}{2})}$

• $\frac{T_{-}}{2}$ = $\frac{1}{2}(\frac{2 \pi}{\omega-}=\frac{2 \pi}{\omega_1-\omega_1}$

• Beat frequency $=(\omega_1-\omega_2)$

$\omega_1=\omega_2$ Beat frequency = 0

Amplitude : Same for both waves

$A \sin \omega_1 t+B \sin \omega_2 t$

• $=\frac{A+B}{2} \sin \omega_1 t+\frac{A-B}{2} \sin \omega_1 t+\frac{A+B}{2} \sin \omega_2 t-\frac{(A-B )}{2} \sin \omega_2 t$

• =$\frac{(A+B)}{2}(\sin \omega_1 t+\sin \omega_2 t)+\frac{(A-B)}{2}(\sin \omega_1 t-\sin \omega_1 t)$

• = $\frac{(A+B)}{2} \sin \omega_{+} t \cos \omega_{-} t-\frac{(A-B)}{2} C_3 \omega_{+} t \sin \omega_{-} t$

Source $\quad V_s$

frequency observed (heard) is different from the frequency emitted by the source

1 Source moving towards the observe

$\lambda=\frac{v}{2} \quad v$ speed of wave

$\lambda^{\prime}=\lambda-v_s T$

$\nu_1= \frac{v}{\lambda^{\prime}}=\frac{v}{\lambda-v_s T}=\frac{v}{\frac{v}{2}-\frac{v_s}{2}}=(\frac{v}{v-v_s}) \nu$

$\nu_1=(\frac{v}{v-v_s}) \nu>\nu$

$\cdots-\cdot-\cdot ?$

$\nu_1=(\frac{v}{v+v_s}) \nu<\nu$

3. When an observer is moving towards the source.

What is the effective distance that the observer sees (feels) between the two maximum?

Let the first maximum be emitted at time $t_1$ and the second one time $t_2$

$t_2-t_1=T.$

Let the observer hear the first maximum at $t_1^\prime$ and $t_2^\prime$

The observer hears the second maximum at $t_2^\prime$

Time period falt by the observe = $t_2^\prime-t_1^\prime$

$t_2^{\prime}-t_1^{\prime}=\frac{\lambda}{v+v_0}$

$T^{\prime}=\frac{1}{\nu^{\prime}}=\frac{\lambda}{(v+v_0)}=\frac{v}{\nu(v+v_0)}$

$\Rightarrow \quad \nu^{\prime}=(\frac{v+v_0}{v}) \nu>\nu$

$\nu^{\prime}=(\frac{v+v_0}{v}) \nu = (\frac{v-v_0}{v}) \nu<\nu$

$2^{\prime}=(\frac{v_{ \pm} v_0}{v \pm v_s}) 2$

$+v_0$ is for observe moving towas the source

$+V_s$ is for source moving away from the observed

$\nu^{\prime}=(\frac{v}{v-v_s}) \nu$

$\nu^{\prime}=(\frac{v}{v-v_s}) \nu \times(\frac{v + v_s}{v})$

$2 h_s^{\prime \prime}=(\frac{v+v_s}{v-v_s}) \nu$

$(\frac{v+v_{\text {wall }}}{v-v_{\text {wall }}})^\nu$

We have conside frequencies of an air column in a pipe - closed at one and or open at both ends

Phenomena of beats

Doppler effect