**Q1.**Two open organ pipes gives 4 beats/sec when sounded together in their fundamental nodes. If the length of the pipe are 100 cm and 102.5 cm respectively, then the velocity of sound is :

Solution

**Q2.**At what speed should a source of sound move so that stationary observer finds the apparent frequency equal to half of the original frequency

Solution

Frequency is decreasing (becomes half), it means source is going away from the observe. In this case frequency observed by the observer is n^'=n(v/(v+v_S ))⇒n/2=n(v/(v+v_S ))⇒v_S=v

Frequency is decreasing (becomes half), it means source is going away from the observe. In this case frequency observed by the observer is n^'=n(v/(v+v_S ))⇒n/2=n(v/(v+v_S ))⇒v_S=v

**Q3.**What is the base frequency if a pipe gives notes of frequencies 425, 255 and 595 and decide whether it is closed at one end or open at both ends

Solution

Let the base frequency be n for closed pipe then notes are n,3n,5n…. ∴ note 3n=255⇒n=85, note 5n=85×5=425 note 7n=7×85=595

Let the base frequency be n for closed pipe then notes are n,3n,5n…. ∴ note 3n=255⇒n=85, note 5n=85×5=425 note 7n=7×85=595

**Q4.**In a sonometer wire, the tension is maintained by suspending a 50.7 kg mass from the free end of the wire. The suspended mass has a volume of 0.0075 m^3. The fundamental frequency of the wire is 260 Hz. If the suspended mass is completely submerged in water, the fundamental frequency will become (take g=10 ms^(-2))

Solution

**Q5.**A uniform rope having mass m hangs vertically from a rigid support. A transverse wave pulse is produced at the lower end. The speed (v) of wave pulse varies with height h from the lower end as shown in figure.

**Q6.**An aeroplane be is above the head of an observer and the sound appears to be coming at an angle of 〖60〗^0 with the vertical. If velocity of sound is v, then the speed of aeroplane is

Solution

**Q7.**In an open organ pipe… wave is present.

Solution

C

C

**Q8.**When beats are produced by two progressive waves of the same amplitude and of nearly the same frequency, the ratio of maximum loudness to the loudness of one of the waves will be n. Where n is

Solution

We know that intensity I∝a^2, where a is amplitude of the wave. The maximum amplitude is the sum of two amplitudes i.e., (a+a=2a) Hence, maximum intensity ∝4 a^2 Therefore the required ratio i.e., ratio of maximum intensity (loudness) and intensity (loudness)of one wave is given by n, n=(4a^2)/a^2 =4

We know that intensity I∝a^2, where a is amplitude of the wave. The maximum amplitude is the sum of two amplitudes i.e., (a+a=2a) Hence, maximum intensity ∝4 a^2 Therefore the required ratio i.e., ratio of maximum intensity (loudness) and intensity (loudness)of one wave is given by n, n=(4a^2)/a^2 =4

**Q9.**Ultrasonic, Infrasonic and audible waves travel through a medium with speeds V_u,V_i and V_a respectively, then

Solution

D

D

**Q10.**A wave travelling in stretched string is described by the equation y=Asin(kx-Ï‰t).The maximum particle velocity is

Solution

Here, y=A sin(kx-Ï‰t) dy/dt=A cos〖(kx-Ï‰t)×(-Ï‰)〗 (dy/dt)_max=A(-1)(-Ï‰)=AÏ‰

Here, y=A sin(kx-Ï‰t) dy/dt=A cos〖(kx-Ï‰t)×(-Ï‰)〗 (dy/dt)_max=A(-1)(-Ï‰)=AÏ‰