Important questions based on NCERT syllabus for Chapter 15 - Waves:
Question-1: A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in a tissue in which the speed of sound is 1.7 km s-1? The operating frequency of the scanner is 4.2 MHz.
Solution: Here speed of sound => υ = 1.7 km s-1 = 1700 ms-1 and
frequency υ= 4.2 MHz = 4.2 x 10^6 Hz
.’. Wavelength, A = υ/V = 1700/(4.2 x 10^6) =4.1 x 10^-4 m.
Question-2: We know that the function y = f (x, t) represents a wave traveling in one direction, where x and t must appear in the combination x + vt or x– vt or i.e. y = f (x ± vt). Is the converse true?
Can the following functions for y possibly represent a travelling wave:
( i ) (x – vt)^2
( ii ) log [( x + vt)/ x0 ]
( iii ) 1 / (x + vt )
Solution: No, the converse is not true, because it is necessary for a wave function representing a traveling wave to have a finite value for all values of x and t. As none of the above functions satisfy the above condition, thus, none represent a traveling wave.
Question-3: Two sitar strings A and B playing the note ‘Ga’ are slightly out of tune and produce beats of frequency 6Hz. The tension in the string A is slightly reduced and the beat frequency is found to reduce to 3Hz. If the original frequency of A is 324 Hz, what is the frequency of B?
Solution: Let υ1 and υ2 be the frequencies of strings A and B respectively.
Then, υ1 = 324 Hz, υ2 = ?
Number of beats, b = 6
υ2 = υ1 ± b = 324 ± 6 !.e., υ2 = 330 Hz or 318 Hz
Since the frequency is directly proportional to square root of tension, on decreasing the tension in the string A, its frequency υ1 will be reduced i.e., number of beats will increase if υ2 = 330 Hz. This is not so because number of beats become 3.
Therefore, it is concluded that the frequency υ2 = 318 Hz. because on reducing the tension in the string A, its frequency may be reduced to 321 Hz, thereby giving 3 beats with υ2 = 318 Hz.