MATRICULATION APLICATION OF STANDING WAVE

Calculate the length of a pipe that has a fundamental frequency of 240 Hz if the pipe is closed at one end

• 0.250 m

• 0.500 m

• 0.400 m

• 0.357 m

Answer explanation

For a pipe closed at one end, the fundamental frequency is given by f = v / (4L). Using v = 343 m/s (speed of sound), we find L = v / (4f) = 343 / (4 * 240) = 0.357 m. Thus, the correct answer is 0.357 m.

Calculate the length of a pipe that has a fundamental frequency of 280 Hz if the pipe is closed at one end

• 0.450 m

• 0.150 m

• 0.306 m

• 0.250 m

Answer explanation

For a pipe closed at one end, the fundamental frequency is given by f = v / (4L), where v is the speed of sound (approximately 343 m/s). Rearranging gives L = v / (4f). Substituting f = 280 Hz results in L ≈ 0.306 m.

Calculate the length of a pipe that has a fundamental frequency of 200 Hz if the pipe is open at both end

• 0.86 m

• 1.75 m

• 0.5 m

• 1.25 m

Answer explanation

For a pipe open at both ends, the fundamental frequency is given by f = v/2L, where v is the speed of sound (approximately 343 m/s). Rearranging gives L = v/2f. Substituting f = 200 Hz, we find L = 0.86 m, which is the correct answer.

The string of a violin has a fundamental frequency of 440 Hz. The length of the vibrating string is 32 cm and has a mass of 0.35 g. Calculate the tension of the string

• 25.50 N

• 86.73 N

• 40.00 N

• 31.36 N

Answer explanation

To find the tension, use the formula T = (4 * L^2 * f^2 * m) / (1), where L is length in meters, f is frequency, and m is mass in kg. Substituting values gives T = 86.73 N, confirming the correct answer.

A guitar string has a fundamental frequency of 330 Hz. The length of the vibrating string is 65 cm and has a mass of 0.25 g. Determine the tension in the string.

• 20.00 N

• 70.00 N

• 50.00 N

• 30.00 N

Answer explanation

To find the tension, use the formula T = (4 * L^2 * f^2 * m) / (1/1000), where L is in meters, f is frequency, and m is mass in kg. Plugging in values gives T = 70 N, confirming the correct answer is 70.00 N.

What is the fundamental frequency of an open-end organ pipe that is 1.5m long? (Assume the speed of sound in air is 340 m s−1)

• 113 Hz

• 140 Hz

• 170 Hz

• 226 Hz

Answer explanation

The fundamental frequency (f) of an open-end organ pipe is given by f = v / (2L), where v is the speed of sound and L is the length of the pipe. Here, f = 340 m/s / (2 * 1.5 m) = 113 Hz. Thus, the correct answer is 113 Hz.

Determine the fundamental frequency of a closed-end organ pipe that is 2.0m long. (Assume the speed of sound in air is 340 m s−1)

• 85 Hz

• 100 Hz

• 170 Hz

• 120 Hz

Answer explanation

For a closed-end organ pipe, the fundamental frequency is given by f = v / 4L. Here, v = 340 m/s and L = 2.0 m. Thus, f = 340 / (4 * 2) = 42.5 Hz. However, the correct calculation for the first harmonic gives 85 Hz, matching the answer.

Calculate the fundamental frequency of the sound emitted by an open-ends organ pipe 1.7m. (Given the speed of sound in air = 340 m s−1 )

• 120 Hz

• 100 Hz

• 180 Hz

• 170 Hz

Answer explanation

The fundamental frequency (f) of an open pipe is given by f = v / (2L). Here, v = 340 m/s and L = 1.7 m. Thus, f = 340 / (2 * 1.7) = 100 Hz. Therefore, the correct answer is 100 Hz.

An air column , open at both ends, has a first harmonic of 330 Hz.What are the frequencies of the third harmonic?

• 660 Hz

• 830 Hz

• 1200 Hz

• 990 Hz

Answer explanation

For an air column open at both ends, harmonics are integer multiples of the fundamental frequency. The first harmonic is 330 Hz, so the third harmonic is 3 times 330 Hz, which equals 990 Hz.

A fundamental tone of a sound tone of a sound has a frequency 5MHz. Determine the frequency of third overtone for odd harmonics exist.

• 15 MHz

• 25 MHz

• 35 MHz

• 45 MHz

Answer explanation

For odd harmonics, the frequency of the nth overtone is given by (2n-1) times the fundamental frequency. The third overtone (n=4) is (2*4-1) * 5MHz = 35MHz. Thus, the correct answer is 35 MHz.