A tuning fork of frequency n is sounded at the open end of a long cylindrical tube having a side opening…

Q: A tuning fork of frequency n is sounded at the open end of a long cylindrical tube having a side opening and fitted with a movable reflecting piston, On moving the piston through 9 cm, the intensity of sound heard by the listener changes from max. to minimum. If speed of sound is 360 m/s, value of n is

(a) 1000 Hz

(b) 250 Hz

(c) 500 Hz

(d) 750 Hz

Ans:(a)

Sol: For change on intensity from maximum to minimum, path diff. = λ/2

λ/2 = (9 + 9)cm,

λ = 36 cm

n = v/λ

= (360×100)/360

= 1000 Hz

A tuning fork of known frequency 256 Hz makes 5 beats per second with the vibrating string of a piano…

Q: A tuning fork of known frequency 256 Hz makes 5 beats per second with the vibrating string of a piano. The beat frequency decreases to 2 beats/sec when the tension in the piano sting is slightly increased. The frequency of the piano string before increasing the tension was

(a) (256+2) Hz

(b) (256-2) Hz

(c) (256-5) Hz

(d) (256+5) Hz

Ans:(c)

Sol: Here, frequency of fork, n1 = 256

No. of beats/sec, m = 5

Possible frequencies of piano string are

n2 = n1 ± m

= 256 ± 5 = 261 or 251

When tension in the piano string is increased, n2 increases and m decreases, therefore, m is negative.

Hence, n2 = 256-5 = 251 Hz

A rocket is receding away from earth with velocity = 0.2 c. The rocket emits signal of frequency 4×10^7 Hz .

Q:A rocket is receding away from earth with velocity = 0.2 c. The rocket emits signal of frequency 4 × 107 Hz . The apparent frequency of the signal produced by the rocket observed by the observer on earth will be

(a) 3 × 106 Hz

(b) 4 × 106 Hz

(c) 2.4 × 107 Hz

(d) 5 × 107 Hz

Ans:(c)

Sol: As , \displaystyle v = \frac{1}{2} \times \frac{\Delta \nu}{\nu} c

\displaystyle 0.2 c = \frac{1}{2} \times \frac{\Delta \nu}{4 \times 10^7} c

\displaystyle \Delta \nu = 1.6 \times 10^7 Hz

As the rocket is receding away

\displaystyle  \Delta \nu = \nu - \nu'

\displaystyle  \nu' = \nu - \Delta \nu

\displaystyle = 4 \times 10^7 - 1.6 \times 10^7

= 2.4 × 107 Hz

A source of frequency n gives 5 beats/s, when sounded with a source of frequency 200 s^-1. The second harmonic…

Q: A source of frequency n gives 5 beats/s, when sounded with a source of frequency 200 s-1. The second harmonic (2 n) gives 10 beats/s, when sounded with a source of frequency 420 s-1. n is equal to

(a) 200 s-1

(b) 205 s-1

(c) 195 s-1

(d) 210 s-1

Ans:(b)

Sol: Here, n = 200 ± 5

and 2 n = 420 ± 10

This is possible only when

n = 200 + 5 = 205

An tuning fork of frequency 480 Hz produces 10 beats per second when sounded with a vibrating sonometer string…

Q: An tuning fork of frequency 480 Hz produces 10 beats per second when sounded with a vibrating sonometer string. What must have been the frequency of the string if a slight increases in tension produces fewer beats per second than before ?

(a) 480 Hz

(b) 490 Hz

(c) 460 Hz

(d) 470 Hz

Ans:(d)

Sol: Frequency of tuning fork, n1 = 480 Hz

No. of beats/sec, m = 10

Frequency of string n2 =(480 ± 10) Hz

A slight increase in tension increases n2

As Number of beats/sec decreases, therefore m must be negative.

n2 = 480-10

= 470 Hz