Question 1) The distance between two consecutive nodes of a standing wave is 15.

Question

Question 1) The distance between two consecutive nodes of a standing wave is 15.4 cm. The hand generating the pulses moves up and down through a complete cycle 2.95 times every 7.35 s. Find the velocity of the wave. Answer in units of m/s.

Question 2) Three successive resonance frequencies in an organ pipe are 2330 Hz, 3262 Hz, and 4194 Hz. What is the fundamental frequency? The speed of sound is 340 m/s. Answer in units of Hz.

Part 2 of this question) What is the length of the pipe? Answer in units of cm.

Question 3) A 765 kg mass is hung from the end of a string with linear density 3.2 g/m around a small frictionless peg. A vibrator is attached at a point near the end of the string. For some values of the vibrator’s frequency and this mass the string resonates with visible standing waves. The vibrating length of the string is 2 m. What is the frequency that will produce the standing wave shown? The acceleration of gravity is 9.8 m/s 2 . Answer in units of Hz.

Explanation / Answer

1)

The distance between two nodes is a half-wavelength.

= 2•(15.4) = 30.8 cm = 0.308 m

ƒ = (2.95 cycles) (7.35 sec) = 0.4014 Hz

V = ĥ
V = (0.4014)•(0.308)
V = 0.124 m/sec

2)

3262 - 2330 = 932 Hz

932 / 2 = 466 Hz

2330 / 466 = 5

3262 / 466 = 7

4194 / 466 = 9

these are odd multiple of 466

tube is closed at one end

f1 = 466 Hz

f1 = v / 4L

L = v / 4f1

L = 340 / (4 x 466)

L = 0.1824 m

3)

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