The frequency of a tuning fork can be found by the method shown. A long tube open at both ends is submerged in a beaker of water, and the vibrating tuning fork is placed near the top of the tube. The length L of the air column is adjusted by moving the tube vertically. The sound waves generated by the fork are reinforced when the length of the air column corresponds to one of the resonant frequencies of the tube. The smallest value for L for which a peak occurs in sound intensity is 5 cm. a) What is the frequency of the tuning fork? The speed of sound in air is 345 m/s. Answer in units of Hz. b) What is the value of L for the second resonant position? Answer in units of cm. c)What is the value of L for the third resonant position? Answer in units of cm.

a. 3.45 kHz b. 5 cm c. 7.5 cm

Explanation:

a. What is the frequency of the tuning fork?

For a tube open at both ends, the frequency, f = nv/2L. The smallest frequency at the smallest length of air column, L = 5 cm = 0.05 m is the fundamental frequency and there, n = 1 and v = speed of sound in air = 345 m/s.

So, f = v/2L

So, f = 345 m/s/(2 × 0.05 m)

= 345 m/s ÷ 0.1 m

= 3450 Hz

= 3.45 kHz

b. What is the value of L for the second resonant position?

The second resonant frequency is f' = 2f

= 2 × 3450 Hz

= 6900 Hz

Since f = nv/2L and f' = second resonant frequency when n = 2.

So, f' = 2v/2L

f' = v/L

L = v/f'

= 345 m/s ÷ 6900 Hz

= 0.05 m

= 5 cm

c. What is the value of L for the third resonant position?

The third resonant frequency is f" = 3f

= 3 × 3450 Hz

= 10350 Hz

Since f = nv/2L and f" = third resonant frequency when n = 3.

So, f" = 3v/2L

2f' = 3v/L

L = 3v/2f'

= 3 × 345 m/s ÷ (2 × 6900 Hz)

= 1035 m/s ÷ 13800 Hz

= 0.075 m

= 7.5 cm

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we need to know the time interval during which the force acts. impulse of force = momentum change