A tuning fork of 512 Hz is used to produce resonance in a resonance tube experiment. The level of water at first resonance…

Q: A tuning fork of 512 Hz is used to produce resonance in a resonance tube experiment. The level of water at first resonance is 30.7 cm and at second resonance is 63.2 cm. the error in calculating velocity of sound is (given speed of sound = 330 m/s)

(a) 204.1 cm/s

(b) 110 cm/s

(c) 58 cm/s

(d) 280 cm/s

Ans: (d)

Sol: $\large \frac{\lambda}{2} = 63.2 – 30.7 $

λ = 0.65 m

Speed of sound observed $\large v= \nu \lambda = 512 \times 0.65$

v = 332.8 m/s

Error in calculating velocity of sound = 332.8 – 330 = 2.8 m/s

= 280 cm/s

An open pipe is in resonance in 2nd harmonic with frequency f1. Now one end of the tube is closed…

Q: An open pipe is in resonance in 2nd harmonic with frequency f1. Now one end of the tube is closed and frequency is increased to f2 such that the resonance again occurs in nth harmonic. Choose that correct option

(a) n = 3, f2 = (3/4) f1

(b) n = 3, f2 = (5/4) f1

(c)n = 5, f2 = (5/4)f1

(d) n = 5, f2 = (3/4)f1

Ans: (c)

Sol: Second harmonic of open pipe , $\large f_1 = 2(\frac{v}{2l}) = \frac{v}{l}$

nth harmonic of closed pipe , $\large f_2 = n(\frac{v}{4l}) $ (here, n is odd)

$\large f_2 > f_1$ (given)

It is possible when n is 5 .

$\large f_2 = 5(\frac{v}{4l}) = \frac{5}{4} f_1$

A closed organ pipe of length L and an open organ pipe contain gases of densities ρ1 and ρ2 respectively…

Q: A closed organ pipe of length L and an open organ pipe contain gases of densities ρ1 and ρ2 respectively. The compressibility of gases are equal in both the pipes. Both the pipes are vibrating in their first overtone with same frequency. The length of the open organ pipe is

(a) L/3

(b) 4L/3

(c)$\large \frac{4L}{3}\sqrt{\frac{\rho_1}{\rho_2}}$

(d) $\large \frac{4L}{3}\sqrt{\frac{\rho_2}{\rho_1}}$

Ans: (c)

Sol:  Since , both the pipes are vibrating in their first overtone with same frequency.

$\large f_c = f_o $

$\large 3 (\frac{v_c}{4l_c}) = 2 (\frac{v_o}{2l_o})$

$\large l_o = \frac{4}{3} (\frac{v_o}{v_c}) l_c$

As , $\large v \propto \sqrt{\frac{1}{\rho}}$

$\large l_o = \frac{4}{3} \sqrt{\frac{\rho_1}{\rho_2}}L$

 

A source of sound of frequency 600 Hz is placed inside water. The speed of sound in water is 1500 m/s…

Q: A source of sound of frequency 600 Hz is placed inside water. The speed of sound in water is 1500 m/s and in air it is 300 m/s. the frequency of sound recorded by an observer who is standing in air is

(a) 200 Hz

(b) 3000 Hz

(c) 120 Hz

(d) 600 Hz

Ans: (d)

Sol: Frequency is source characteristic . It is independent of medium .

In the experiment of the determination of the speed of sound in air using the resonance column method…

Q: In the experiment of the determination of the speed of sound in air using the resonance column method, the length of the air column that resonates in the fundamental mode, with a tuning fork is 0.1 m. when this length is changed to 0.35 m, the same tuning fork resonates with the first overtone. Calculate the end correction.

(a) 0.012 m

(b) 0.025 m

(c) 0.05 m

(d) 0.024 m

Ans: (b)

Sol: Let Δl be the end correction .

A/c to question ,

$\large \frac{v}{4(0.1+\Delta l)} = 3 (\frac{v}{4(0.35+ \Delta l)})$

On solving , Δl = 0.025 m