Wave Motion and String Waves Concept Page - 15

Example

Understand and calculate acceleration of a particle of standing wave at a fixed position

Example: A transverse wave along a string is given by  y=2sin(2π(3t−x)+π4)$y=2{\displaystyle \sin (2\pi (3t-x)+\frac{\pi }{4})}$, where  x  and  y  are in cm and t is in second. Find the acceleration of a particle located at  X=4cm   at t=1 sec.

Solution:
y=2 sin(6πt−2πx+π4)$y=2sin(6\pi t-2\pi x+\frac{\pi }{4})$
vp=dydt=12π cos(6πt−2πx+π4)$v_p=\frac{dy}{dt}=12\pi cos(6\pi t-2\pi x+\frac{\pi }{4})$

ap=dvpdt$a_p=\frac{d{v}_p}{dt}$
     =−72π2 sin(6πt−2πx+π4)$=-72{\pi }^{2}sin(6\pi t-2\pi x+\frac{\pi }{4})$
At x=4, t=1$x=4,t=1$
ap=−72π2 sin(6π−8π+π4)$a_p=-72{\pi }^{2}sin(6\pi -8\pi +\frac{\pi }{4})$
     =−72π2×12√$=-72{\pi }^2\times \frac{1}{\sqrt{2}}$
     =−362√π2 cm/sec2$=-36\sqrt{2}{\pi }^{2}cm/se{c}^2$

Example

Apply boundary conditions on a standing wave in a string fixed at one end and hence find possible wavelengths

Depending on the shape of the string being formed possible wavelengths are:
λ=2Ln+1/2$\lambda ={\displaystyle \frac{2L}{n+1/2}}$ for n = 0, 1, 2, 3... different values of wavelength occur.

Definition

String fixed at one end

A travelling wave, at a rigid boundary or a closed end, is reflected with a phase reversal.Let the incident wave be yi(x,t)=asin(kx−ωt)$y_i(x,t)=asin(kx-\omega t)$. 
Then the reflected wave is:  yr(x,t)=−asin(kx+ωt)$y_r(x,t)=-a\sin (kx+\omega t)$.
The principle of superposition gives the combined wave: y(x,t)=(2asinkx)cosωt$y(x,t)=(2a\sin kx)\cos \omega t$.
A node will always form at the fixed end while an antinode will always form at the free end. The simplest standing wave that can form under these circumstances is one-quarter wavelength long.
Possible wavelengths: λ=2n+1/2L$\lambda ={\displaystyle \frac{2}{n+1/2}}L$

Example

Find the nodes of a given sinusoidal standing wave on a string fixed at both ends

Example: The displacement vibration of a string of length 60cm fixed at both ends are represented by Y=4sin(πx20)cos96πt${\displaystyle \mathrm{Y}=4\sin ({\displaystyle \frac{\pi \mathrm{x}}{20}})\cos 96\pi t}$. Find the location of nodes excluding the ends of string.

Solution:
λ=2πK=2ππ20=40 cm$\lambda ={\displaystyle \frac{2\pi }{K}}={\displaystyle \frac{2\pi }{{\displaystyle \frac{\pi }{20}}}}=40cm$
∴$\therefore$ nodes are separated by λ2${\displaystyle \frac{\lambda }{2}}$
⇒$\Rightarrow$ positions are 20 cm, 40 cm

Example

Find the antinodes of a given sinusoidal standing wave on a string fixed at both ends

Example: The length of a sonometer wire is 90 cm and the stationary wave setup in the wire is represented by an equation y=6sin(πx30)cos(250πt)$y=6\sin ({\displaystyle \frac{\pi x}{30}})\cos (250\pi t)$ where x$x$, y$y$ are in cm$cm$ and t$t$ is in second$second$. Find the distances of successive antinodes from one end of the wire.

Solution:
l=90 cm$l=90cm$
y=6 sin(πx30)cos(250 πt)$y=6\sin \left({\displaystyle \frac{\pi x}{30}}\right)\cos (250\pi t)$
So, k=π30$k={\displaystyle \frac{\pi }{30}}$         k=2πλ$k={\displaystyle \frac{2\pi }{\lambda }}$
So, λ=2πk$\lambda ={\displaystyle \frac{2\pi }{k}}$
λ=2ππ/30=60 cm$\lambda ={\displaystyle \frac{2\pi }{{}^{\pi }{/}_{30}}}=60cm$
So, antinodes will be formed at
=λ4,3λ4,5λ4$={\displaystyle \frac{\lambda }{4}},{\displaystyle \frac{3\lambda }{4}},{\displaystyle \frac{5\lambda }{4}}$
=604,3×604,5×604$={\displaystyle \frac{60}{4}},3\times {\displaystyle \frac{60}{4}},5\times {\displaystyle \frac{60}{4}}$
=15,45,75 cm$=15,45,75cm$

Definition

Fiind higher harmonics and overtones for standing wave in a string fixed at both ends

The natural frequencies for the normal modes of oscillations of the system is:
ν=nv2L$\nu ={\displaystyle \frac{nv}{2L}}$

The lowest possible natural frequency of a system is called its fundamental mode or first harmonic. For n = 2,3,4..
Frequency is called 2nd and 3rd harmonic respectively.

In the case of a vibrating string, the frequency of the first overtone is equal to the frequency of which harmonic?

The overtone frequency is
ν1=2v2L=2νo${\nu }_1={\displaystyle \frac{2v}{2L}}=2{\nu }_o$
2nνo=2νo$2n{\nu }_o=2{\nu }_o$
⟹n=1$⟹n=1$
n=0$n=0$ is the first harmonic,   ∴n=1$\therefore n=1$ is the second harmonic.

Example

Use the relation between amplitude of displacement and amplitude of pressure variation in sound waves

Example: Figure shown is a graph, at a certain time t, of the displacement function S(x,t) of three sound waves 1,2 and 3 as marked on the curves that travel along x-axis through air. If P1,P2$P_1,P_2$ and P3$P_3$ represent their pressure amplitudes respectively, then find the correct relation between pressure and the wavelength.

Solution:
The relation between Amplitude of displacement and amplitude of Pressure is given by:
P0=B.K.S0=B(2πλ)S0⇒ P0∝1λ$P_0=B.K.S_0=B\left({\displaystyle \frac{2\pi }{\lambda }}\right)S_0\Rightarrow P_0\propto {\displaystyle \frac{1}{\lambda }}$
Thus, pressure amplitude is highest for minimum wavelength, other parameters B$B$ and S0$S_0$ being same for all.

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