Q: An audio transmitter (T) and a receiver (R) are hung vertically from two identical massless strings of length 8 m with their pivots well separated along the X axis. They are pulled from the equilibriu

aligned & _0=1- _0^2 2 =0.9 \\ & _0^2 2 =0.1 _0=10.2= 1 5 aligned aligned & f_ = v+v^ v-v^ f \\ & f_ = v-v^ v+v^ f \\ & f_ =f_ -f_ = v+v^ v-v^ f- v-v^ v+v^ f \\ & = (v+v^ )^2- (v-v^ )^2 v^2-v^ 2 f aligned f_ = 4 v v^ v^2-v^ 2 f ....(i) Here, v ^ = _ = _0 ( = angular frequency ) aligned & = _0 ~g \\ & v ^ = _0 ~g \\ & v ^ = 1 5 10 8 \\ & v ^ =4 aligned Put in equation (i) aligned & f _ = 4 330 4 660 330^2-4^2 \\ & 16 330 660 330 32 aligned

Q: A string of length 1 m and mass 2 × 10 − 5 kg is under tension T . When the string vibrates, two successive harmonics are found to occur at frequencies 750 Hz and

Given: l = 1 m , m = 2 × 10 – 5 kg . As successive harmonics are being given, so it is the case of both ends fixed. Now, f n + 1 - f n = 1000 - 750 ⇒ n + 1 2 l T μ - n 2 l T μ = 250 ⇒ 1 2 l T μ = 250 ⇒ T 2 × 10 - 5 = 250 × 2 × 1 ⇒ T 2 × 10 - 5 = 25 × 10 4 ⇒ T = 50 × 10 - 1 ⇒ T = 5 N

Q: S 1 and S 2 are two identical sound sources of frequency 656 Hz . The source S 1 is located at O and S 2 moves anti-clockwise with a uniform speed 4 2

Given: Source frequency, f 0 = 656 Hz . Velocity of sound = 324 m s - 1 . Velocity component of the source moving away from detector along like P Q , V s = 4 2 cos 45 o = 4 m s - 1 . Therefore, apparent frequency, f = v v + v s f 0 = 324 324 + 4 656 ⇒ f = 648 Hz

Q: S 1 and S 2 are two identical sound sources of frequency 656 Hz . The source S 1 is located at O and S 2 moves anti-clockwise with a uniform speed 4 2

Given: f 0 = 656 Hz v = 324 m s - 1 Frequency heard due to movement of ( S 1 ) f 1 = v v - u s f 0 ⇒ f 1 = 324 324 - 4 × 656 = 324 320 × 656 And frequency heard due to movement of ( S 2 ) would be same as that of f 0 , because the direction of velocity at position R is perpendicular to the line P R , so there is no relative change in position at that moment. f 2 = 656 Hz ∴ Beat frequency &#871

Question 1

List-I shows four configurations made of straight and semi-circular narrow tubes containing air. A sound wave of wavelength = 0.29 m enters these structures at the point S and a sound detector is plac

Accepted Answer

For maximum amplitude at the detector D, the path difference x between the two paths must be an integer multiple of the wavelength . For the smallest value of l, we take x = = 0.29 m. For configuration (P): The sound travels through a straight tube of length l and a semi-circular tube of diameter l. Path difference x = l 2 - l = l ( 2 - 1 ). Equating to : l ( 3.1416 2 - 1 ) = 0.29 l(0.5708) = 0.29 l 0.51 m. Thus, P 3. For configuration (Q): The sound travels through a straight tube of length l a

Question 2

Consider a system of three connected strings, S_1, S_2 and S_3 with uniform linear mass densities kg / m , 4 ~kg / m and 16 ~kg / m , respectively, as shown in the figure. S_1 and S_2 are connected at

Accepted Answer

(A) y_1=y_0 ( t-k x) when wave going from Rarer to Denser, aligned & y_r=A_r ( t+k x+ ) \ & y_r=a_1 y_0 ( t+k x+ ) aligned option (A) correct (B) For transmitted from point P aligned & y _ t = A _ t [ t - k _1 x ] \ & k _1 k = _1 = k _1 k = 4 \ & k _1=2 k \ & y _ t = a _2 y _0 [ t -2 kx ] aligned option (B) incorrect (C) when reflected from Q aligned & y_i=a_2 y_0 [ t-2 k x] \ & y_r=a_3 y_0 [ t+2 k x+ ] aligned option (C) incorrect (D) when transmitted from Q aligned & y_ t =a_4 y_0 [ t =k_

Question 3

An audio transmitter (T) and a receiver (R) are hung vertically from two identical massless strings of length 8 m with their pivots well separated along the X axis. They are pulled from the equilibriu

Accepted Answer

aligned & _0=1- _0^2 2 =0.9 \ & _0^2 2 =0.1 _0=10.2= 1 5 aligned aligned & f_ = v+v^ v-v^ f \ & f_ = v-v^ v+v^ f \ & f_ =f_ -f_ = v+v^ v-v^ f- v-v^ v+v^ f \ & = (v+v^ )^2- (v-v^ )^2 v^2-v^ 2 f aligned f_ = 4 v v^ v^2-v^ 2 f ....(i) Here, v ^ = _ = _0 ( = angular frequency ) aligned & = _0 ~g \ & v ^ = _0 ~g \ & v ^ = 1 5 10 8 \ & v ^ =4 aligned Put in equation (i) aligned & f _ = 4 330 4 660 330^2-4^2 \ & 16 330 660 330 32 aligned

Question 4

Two uniform strings of mass per unit length and 4 , and length L and 2 L, respectively, are joined at point O , and tied at two fixed ends P and Q , as shown in the figure. The strings are under a uni

Accepted Answer

C _1= T , C _2= T 4 = C _1 2 For node at O : L = n _1 2 , 2 ~L = m _2 2 ( n , m are integers) aligned & _1= 2 ~L n , _2= 4 ~L ~m \ & C _1 _1 = C _2 _2 \ & C _1 2 ~L n = C _1 2 4 ~L m \ & 4 n = m aligned For minimum frequency, n =1, ~m =4 v_ = C _1 1 2 ~L = 1 2 ~L T =v_0 The string will look like Total no. of nodes =6 including the end nodes For antinode at O : aligned & L =(2 n +1) _1 4 ; 2 ~L =(2 n +1) _2 4 (n, m are integers) \ & _1= 4 ~L (2 n +1) ; _2= 8 ~L (2 ~m +1) \ & C _1 _1 = C _2 _

Question 5

A source (S) of sound has frequency 240 ~Hz . When the observer ( O ) and the source move towards each other at a speed v with respect to the ground (as shown in Case 1 in the figure), the observer me

Accepted Answer

Frequency received by observer f_0= ( C V_0 C V_S ) f_ s , C is speed of sound Case-1: aligned & f _1= ( C + V C - V ) f _ s \ & 288= ( C + V C - V ) 240 aligned Case-2: f_2= ( C-V C+V ) f_S n = ( C - V C + V ) 240 multiply the two equations, we get. aligned & (288)( n )=(240)(240) \ & n =200 aligned

Question 6

An ideal gas is in thermodynamic equilibrium. The number of degrees of freedom of a molecule of the gas is n . The internal energy of one mole of the gas is U n  and the speed of sound in the gas

Accepted Answer

As we know, internal energy of one mole of the gas  U n = n R T 2 . Clearly,  U n &#8733; n , where  n  is degree of freedom. Therefore, as  n  increases,  U n  also increases. Now, speed of sound in gas v n = &#947; R T M = 1 + 2 n R T M Clearly, as  n  increases and  v n  decreases. Hence, U 7 &#62; U 5  and  U 7 &#62; U 6 and  v 5 &#62; v 7 .

Question 7

A string of length 1 &#160; m and mass 2 &#215; 10 &#8722; 5 &#160; kg  is under tension T . When the string vibrates, two successive harmonics are found to occur at frequencies 750 &#160; Hz and

Accepted Answer

Given:  l = 1 &#160; m , &#160; m = 2 &#215; 10 &#8211; 5 &#160; kg . As successive harmonics are being given, so it is the case of both ends fixed. Now, f n + 1 - f n = 1000 - 750 &#8658; n + 1 2 l T &#956; - n 2 l T &#956; = 250 &#8658; 1 2 l T &#956; = 250 &#8658; T 2 &#215; 10 - 5 = 250 &#215; 2 &#215; 1 &#8658; T 2 &#215; 10 - 5 = 25 &#215; 10 4 &#8658; T = 50 &#215; 10 - 1 &#8658; T = 5 &#160; N

Question 8

S 1  and  S 2  are two identical sound sources of frequency 656 &#160; Hz . The source  S 1  is located at O and S 2 moves anti-clockwise with a uniform speed  4 2 &#160;

Accepted Answer

Given: Source frequency,  f 0 = 656 &#160; Hz . Velocity of sound = 324 &#160; m &#160; s - 1 . Velocity component of the source moving away from detector along like  P Q , V s = 4 2 cos 45 o = 4 &#160; m &#160; s - 1 . Therefore, apparent frequency,  f = v v + v s f 0 = 324 324 + 4 656 &#8658; f = 648 &#160; Hz

Question 9

S 1  and  S 2  are two identical sound sources of frequency 656 &#160; Hz . The source  S 1  is located at O and S 2 moves anti-clockwise with a uniform speed  4 2 &#160;

Accepted Answer

Given:  f 0 = 656 &#160; Hz v = 324 &#160; m &#160; s - 1 Frequency heard due to movement of ( S 1 ) f 1 = v v - u s f 0 &#8658; f 1 = 324 324 - 4 &#215; 656 = 324 320 &#215; 656 And frequency heard due to movement of ( S 2 )  would be same as that of  f 0 , because the direction of velocity at position  R  is perpendicular to the line  P R , so there is no relative change in position at that moment.  f 2 = 656 &#160; Hz &#8756;  Beat frequency  &#871

Question 10

A source, approaching with speed  u  towards the open end of a stationary pipe of length  L , is emitting a sound of frequency  f s . The farther end of the pipe is closed. The spe

Accepted Answer

Consider a tuning fork as a source of sound which is moving towards the stationary pipe with velocity  u . According to the Doppler effect , f ' = f s v - v o v - v s Where, v = is speed of sound v s = is velocity of source v o = is velocity of observer f ' = is appeared frequency f s = is actual frequency So the appeared frequency of the sound will be,   f ' = f s v - 0 v - u Since pipe is closed at one end so it will be like a close organ pipe and we know that close organ

Question 11

A stationary tuning fork is in resonance with an air column in a pipe. If the tuning fork is moved with a speed of 2 &#160; m &#160; s - 1 in front of the open end of the pipe and parallel to it, the

Accepted Answer

f = v 4 L 1 f 320 + 0 320 - 2 = v 4 L 2 &#8658; L 2 L 1 = 318 320 &#8658; L 1 - L 2 L 1 = 1 - 318 320 = 2 320 &#916; L L &#215; 100 % = 20 32 % = 0 . 62 %

Waves and Sound — Previous Year Questions & Solutions