Class 11th

This is a short answer type question as classified in NCERT Exemplar

Given frequency of the wave v=256Hz

          T= $\frac{1}{v}=\frac{1}{256}=3.9*{10}^{3}s$

(a) time taken to pass through mean position is

t=T/4=1/40 =3.9 $*\frac{{10}^{-3}}{4}s=$ 9.8 $*$ 10^-4s

(b) nodes are A, B, C, D, E (having zero displacement)

antinodes are A' and C' (having maximum displacement)

(c) it is clear from diagram A' and C' are forming antinodes have $\lambda$ separation= v/v'=360/256=1.41m

This is a short answer type question as classified in NCERT Exemplar

We have to observe the displacement and position of different points, then accordingly nature of two waves decided

Points on positions x= 10,20,30,40 never move, always at mean position with respect to time . these are forming nodes which forms a stationary wave

Distance between two successive nodes = $\frac{\lambda }{2}$

$\lambda =2($ node to node distance)

=2 (20-10)

= 2 (10)=20cm

This is a short answer type question as classified in NCERT Exemplar

As the source is moving towards the observer hence apparent frequency observed is more than the natural frequency

Frequency of whistle $v=400Hz$

Speed of train = v_t= 10m/s

Velocity of sound in air =v= 330m/s

Apparent frequency when source is moving v_app= (v/v-v_t)v

= ( $\frac{330}{330-10})400$

= $\frac{330}{320}*400=412.5Hz$

This is a short answer type question as classified in NCERT Exemplar

length of pipe

L=20cm =20 $*{10}^{-2}m$

$v_{close}=\frac{v}{4L}=\frac{330}{4*20*{10}^{-2}}$

$v_{close}=330*\frac{100}{80}=412.5Hz$

$\frac{v_{given}}{v_{closed}}=1237.5/412.5$ =3

Hence 3^rd harmonic node of the pipe is resonantly excited by the source of the given frequency.

This is a short answer type question as classified in NCERT Exemplar

Given that length of wire l=12m

Mass of the wire m=2.10kg

T= 2.06 $*{10}^{4}N$

Speed of transverse wave v= $\sqrt[]{\frac{T}{\mu }}$ = $\sqrt[]{\frac{2.06*{10}^4}{2.1/12}}=\sqrt[]{\frac{2.06*12*{10}^4}{2.1}}=343m/s$

This is a short answer type question as classified in NCERT Exemplar

Let n₁>n₂ beat frequency $v_b=n_1-n_2$

As we know time per beats = 1/frequency

Time period beats = T_b= 1/ $v_b=1/n_1-n_2$

This is a short answer type question as classified in NCERT Exemplar

we know the speed of air  v $\propto \sqrt[]{T}$

$\frac{v_t}{v_o}=\sqrt[]{\frac{T_T}{T_o}}=\sqrt[]{\frac{T_T}{273}}$

$\frac{V_T}{V_o}=\frac{3}{1}$

3/1 = $\sqrt[]{\frac{T_T}{T_o}}=\frac{T_T}{273}=9$

T_T=273 $*9=2457K$

=2457-273=2184 ^oC

This is a short answer type question as classified in NCERT Exemplar

Frequency of vibrations produced by a stretched wire

$v=\frac{n}{2L}\sqrt[]{\frac{T}{\mu }}$

Mass per unit length = mass/length = $\frac{\pi r^{2}l\rho }{l}=\pi r^2\rho$

$v=\frac{n}{2L}\sqrt[]{\frac{T}{\pi r^2\rho }}=v=\sqrt[]{\frac{1}{r^2}}$

$v\propto \frac{1}{r}$ so when radius is tripled v will be 1/3 rd of previous value.

This is a short answer type question as classified in NCERT Exemplar

Displacement of an elastic wave y =3sinwt+4coswt

3= acos $\varnothing$

4=asin $\varnothing$

On dividing above equation

tan $\varnothing$ =4/3

$\varnothing ={tan}^{-}\left(\frac{4}{3}\right)$

a²cos^{2 $\varnothing$} +a²sin^{2 $\varnothing$} = 3²+4²

a² (cos^{2 $\varnothing +{sin}^2\varnothing$} )=25

a².1=25

a=5

Y= 5cos $\varnothing sinwt$ +5sin $\varnothing coswt$

= 5 [cos $\varnothing sinwt+sin\varnothing coswt$ ]=5sin (wt+ $\varnothing$ )

$\varnothing ={tan}^{-1}\left(\frac{4}{3}\right)$

Hence amplitude =5cm

This is a short answer type question as classified in NCERT Exemplar

Frequency of tuning fork A

$v_A=512$ Hz

Probable frequency of tuning fork B

${\mathrm{v}}_{\mathrm{B}}={\mathrm{v}}_{\mathrm{A}}+5=512\pm 5=517$ Hz or 507Hz

When B is loaded its frequency reduces .

If it is 517Hz it might reduced to 507Hz given again a beat of 5Hz

If it 507Hz reduction will always increase the beat frequency, hence $v_B=517$ Hz