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1. Transverse vibrations are produced which are gradually damped. 

2. A stretched string of a guitar. 

3. As the energy of wave is dissipated its amplitude decreases. 

4. The body will stop vibrating. It will come to rest.

(i) Damped vibrations

(ii) Example: When a slim branch of a tree is pulled and then released, it makes damped vibrations.

(iii) The amplitude of vibrations gradually decreases due to the frictional (or resistive) force which the surrounding medium exerts on the body vibrating in it. As a result, the vibrating body continuously loses energy in doing work against the force of friction causing a decrease in its amplitude.

(iv) After sometime, the vibrating body loses all of its energy and stops vibrating.

C. a man 

The voice of man has the least frequency.

Correct Answer is: (a) y = Asin(nπx/L) cos ωt , (b) y = Asin(nπx/L) sin ωt

y = 0 at x = 0. This can be satisfied by the term sin (nπx/L)

Correct Answer is: (b) 0.1 s 

Wave velocity on string = V = √(1.6/(10^-2/0.4)) =8 /s.m

For the string to regain its shape, the pulse must travel a distance which is twice the length of the string, as the pulse gets inverted at each reflection from a fixed end.

∴ t = 2 x 0.4 m / 8 m/s = 0.1 s.

Correct Answer is: (d) 240 Hz

Initially, 260 = √T₁/m, T₁ = 50.7 g = 507 N.

When the mass is submerged, upthrust 

= (0.0075 m³ )(10³ kg/m³)(10 m/s²)

= 75 N.

New tension = T₂ = (507 - 75) N = 432 N.

n = 1/2l √T₂/m

or n/260 = √T₂/T₁ = √432/507 = √144/169 =12/13 

or n = 240 Hz.

Correct Answer is: (d) 20%

1. When two pupils talk to each other, the sounds of two frequencies produced can be heard clearly.

2. When all the children talk to one another sounds of many frequencies produced can not be heard clearly.

3. Sound level will be more creating loud noise.

B) Because of high pitch sound

(C) square root of adiabatic elasticity.

1. Echo of sound depends upon the temperature of the surrounding and distance between source and reflecting surface.

2. To hear a distinct echo at 22 °C, the minimum distance required between the source of sound and reflecting surface should be 17.2 metre.

3. The velocity of sound depends on the temperature of air. Thus, the minimum distance will change with temperature. 

Hence, we cannot hear an echo at every place.

Correct option is: (B) 1 : 4

Correct option is: (B) decreases

Correct option is: (C) 340 m/s

Given: T₁ = 27 °C = 27 + 273 = 300 K,

v₁ = 340 m/s,

T₂ = 127 °C = 127 + 273 = 400 K

To find: Velocity (v₂)

Formula: \(\frac{v_1}{v_2}=\sqrt{\frac{T_1}{T_2}}\)

Calculation: From formula,

v₂ = v₁ \(\sqrt{\frac{T_2}{T_1}}\) = 340, \(\sqrt{\frac{400}{300}}\)

= 340 × 1.1547

∴ v₂ = 392.6 m/s

Correct option is: (B) 819 °C

Given: \(\frac{p}{p_0}\) = 1.5, T₀ = 0 °C = 273 K

To find: Temperature (T)

Formula: \(\frac{v}{v_0}=\sqrt{\frac{T}{T_0}}\)

Calculation: From formula,

\(\frac{T}{T_0}=(\frac{v}{v_0})^2\)

∴ T = T₀ \((\frac{v}{v_0})^2\)

∴ T = 273 (1.5)² = 614.25 K = 341.25 °C

(A) For 1 °C rise in temperature, velocity of sound increases by 0.61 m/s.

The answer is (D) of a single frequency

Sound waves cannot travel through a vacuum.

A sound of a single frequency is called a tone.

If the frequency of a sound wave is 512 Hz, the number of rarefactions produced per second is 512.

The answer is (D) All the above

Hertz (Hz) is S.I. unit of a sound wave.

Sound waves are mechanical waves (longitudinal waves).

Laws of reflection of sound:

(1)The direction in which the sound is incident and reflected, make equal angles with the normal to the reflecting surface.

(2) The direction in which the sound is incident, the direction in which sound is reflected and the normal to the reflecting surface, all lie in the same plane.

Activity: Take a drawing board and fix it in the vertical position on the table. Put two metallic cardboard tubes A and B. These tubes make some angle with each other. Put a clock near one end of the tube A and a solid screen between the two tubes, so that the sound of the clock may not be heard directly by the ear placed in front of the second tube.

The sound waves after reflecting from the drawing board enter the second tube and are heard by the ear placed in front of the second tube. Go on adjusting the position of the second tube B, till the loudness of the sound heard through the second tube becomes maximum. At this stage, the incident angle which is the law of reflection.

The sound frequencies greater than 20000 Hz are called ultrasonic waves or ultrasound. Human beings cannot hear ultrasound. Dogs, bats and dolphins can hear ultrasound. Due to very high frequency, ultrasound has greater penetrating power than ordinary sound.

Detection of defects in metal blocks: In this process, ultrasound waves are allowed to pass through the metal block and detectors are used to detect the transmitted waves. Even if there is a small defect, the ultrasound gets reflected back, indicates the presence of the flaws or defect.

Cleaning of spiral tubes: Ultrasonic waves are used for cleaning spiral tubes. The object to be cleaned is kept in the cleaning solution and the solution is subjected to the ultrasonic waves. The high-frequency waves stir up the dust/dirt particles. These particles get detached and the spiral tubes get thoroughly cleaned.

5 vibrations by pendulum in 1 sec

So 8 vibrations in 8/5 seconds = 1.6 sec

Velocity = 2 × D/ time

340 = 2 × D/ 1.6

D = 340 × 1.6 / 2 = 272 m

Given, 

distance travelled = 840 m 

time taken = 2.5 sec 

length of wave = 70 cm = 0.7 m

Velocity of Wave = \(\frac{distance}{time}\)= \(\frac{840}{2.5}\) = 336ms^-1

Frequency = \(\frac{velocity}{wavelength}\) = \(\frac{336}{0.7}\) = 480 Hz