A tuning fork with frequency 800 hz

If you have two tuning forks and you know that one tuning fork is 305 Hz, and when the tuning forks are pinged together you get a beat frequency of 6 Hz, then you know that the other tuning fork is either 311 Hz or 299 Hz. This idea I am fine with. Now, the second part is as follows: if you replace the known tuning fork with one that emits a higher frequency, the beat frequency is 4 Hz. What is the frequency of the second tuning fork? The answer is 311 Hz. But I don't understand how we can arrive there. The two things we know are $1.$ that the new frequency is $>$305 Hz and $2.$ that the beat frequency decreases from 6 Hz to 4 Hz. How does this then tie together that the other frequency is definitely 311 Hz? The way it was explained was that there are two situations: if the other frequency was 306 Hz (why? where did 306 Hz come from?) then the beat frequency would increase and if the other frequency was 311 Hz (again why?) then the beat frequency would decrease. Since we know the beat frequency decreases from 6 Hz to 4 Hz, it must be the second case. Well, you start with tuning fork A at 305 Hz and you know that B is either 299 Hz or 311 Hz. Now you take C, which is tuned at >305 Hz, which means at least 306 Hz (in integer steps...). If B was 299 Hz, then, together with C, which is higher tuned than A, the Beat frequency would increase to at least 306 Hz - 299 Hz = 7 Hz. But the beat frequency decreases, therefore, B must be 311 Hz. Thanks for contributing an answer to Phy...

1) find the wavelength of the sound of the tuning fork (cm) 2) what's the vertical distance x from the top of the pipe to the antinode above the pipe (cm) 3) find the frequency of the tuning fork (Hz) Homework Equations velocity = wavelength . frequency for pipe closed at one end n = 1,3,5 wavelength for n times harmonic = 4.lenght/n The Attempt at a Solution I don't know for which harmonic the tuning fork exciting at 16.7 cm and 50.7 cm I assume for 16.7 is 1st harmonic so the wavelength = 4*16.7 = 66.8 cm since the 50.7 approximately 3 times 16.7, I assume that the n value is 3, which is the 2nd harmonic so the wavelength = 4*50.7/3 = 67.6 cm so the wavelength is 67 cm?? 2) can you give me clue, where to start? 3) f = v / wavelength f = 340 / 0.67 m = 507 Hz You seem to have done the hard parts OK. For (2) I think you only need to know the meaning of 'antinode'. Fix stuff in your head - don't leave this problem without sketching a picture of the amplitude against height on top of these pictures, which would also help answer. I guess I answered this question because I did that same experiment in English 'A' Level exam many years ago mnyah mnyah. (I got it a bit wrong because I forgot the /4 , but I did write my answer was unreasonable so probably got nearly full marks.) You seem to have done the hard parts OK. For (2) I think you only need to know the meaning of 'antinode'. Fix stuff in your head - don't leave this problem without sketching a picture of the amplitude agai...

This problem has been solved! You'll get a detailed solution from a subject matter expert that helps you learn core concepts. See Answer See Answer See Answer done loading Question:1.80 m is 0.500 kg A string is connected to a tuning fork whose frequency is 80.0 Hz and is held under tension by 0.500 kg. The tuning fork causes the string to vibrate as shown. The mass per unit length for the string is 3.85 . 10-2 kg/m. O 6.00 10 kg/m. O 6,80 10-3kg/m. 4.34 kg/m O 9.45 10-4 kg/m. 1.80 m is 0.500 kg A string is connected to a tuning fork whose frequency is 80.0 Hz and is held under tension by 0.500 kg. The tuning fork causes the string to vibrate as shown. The mass per unit length for the string is 3.85 . 10-2 kg/m. O 6.00 10 kg/m. O 6,80 10-3kg/m. 4.34 kg/m O 9.45 10-4 kg/m. Previous question Next question

Vibrational Modes of a Tuning Fork Acoustics and Vibration Animations Daniel A. Russell, Graduate Program in Acoustics, The Pennsylvania State University This work by Based on a work at The content of this page was originally posted on January 24, 2012. The HTML code was modified to be HTML5 compliant on March 16, 2013. Fundamental Mode (426 Hz) The fundamental mode of vibration is the mode most commonly associated with tuning forks; it is the mode shape whose frequency is printed on the fork, which in this case is 426 Hz. The two tines of the fork alternately move toward and away from each other, each bending like a cantilever beam, fixed at the stem and free at the other end. This is a symmetric mode, since the two tines are mirror images of each other. When vibrating in the fundamental mode, it would appear that the stem of the fork is stationary. However, the stem actually vibrates up and down at the fundamental frequency as well as at the second harmonic, 852 Hz - twice the frequency of the fundamental (even there is no vibrational mode of the fork at this frequency). This stem motion is very small, and difficult to feel if you place a finger tip at the bottom of the stem. But, it can be effectively demonstrated by touching the stem of a vibrating fork to a table top, door, or piano soundboard. The fundamental vibration mode of a tuning fork radiates sound as a longitudinal (or linear) quadrupole sound source [1] with a well-defined transition between the near-field a...

NEET UG 2019 Odisha 3 A tuning fork with frequency 800 Hz produces resonance in a resonance column tube with upper end open and lower end closed by water surface Successive resonance are observed at length 9 75 cm 31 25 cm and 52 75 cm The speed of sound in air is 1 500 m s 3 344 m s 2 156 m s 4 172 m s

• Music • Music (CDs) • Music (MP3 Downloads) • Exclusive MP3 Downloads • Sonic Tools • Tuning Forks & Sonic Tools • Chakra Tuner App • Wholesale Enquiries • Books • Forbidden Frequencies eBook • Books • Events & Training • Tibetan Secrets of Happiness • Jonathan Goldman’s Complete Chakra Chanting Experience • The 11:11 Divine Name Seminar • Healing Sounds Correspondence Course • Full Tuition Payments • Sound Healing Info • Healing Sounds Blog • Sound Healing Discussion • Healing Sounds Releases • Articles • Video • Uncategorized • Articles and Interviews • Healing Sounds Videos • Vibratory Research • Healing Sounds Radio Shows • Sound Healers Association • World Sound Healing Day • Bibliography • Links • Events and Training • The Power of Sound Healing Online Course • Tibetan Secrets of Happiness • Jonathan Goldman’s Complete Chakra Chanting Experience • The 11:11 Divine Name Seminar • Healing Sounds Correspondence Course • Workshops by Jonathan & Andi Goldman • Current Events • Contact Speaking of the resonance of our beloved planet, Jonathan has created two different sets of tuning forks which utilize the Schumann Resonance. This resonance, which was discovered and then validated over 60 years ago, is based upon the modulating frequency of the ionosphere of our planet. It is considered the frequency of the electro-magnetic field of the planet. While it modulates, depending upon conditions such as lightning and sun spots, it’s main frequency seems to be 7.83 Hz. Utilizin...

• Music • Music (CDs) • Music (MP3 Downloads) • Exclusive MP3 Downloads • Sonic Tools • Tuning Forks & Sonic Tools • Chakra Tuner App • Wholesale Enquiries • Books • Forbidden Frequencies eBook • Books • Events & Training • Tibetan Secrets of Happiness • Jonathan Goldman’s Complete Chakra Chanting Experience • The 11:11 Divine Name Seminar • Healing Sounds Correspondence Course • Full Tuition Payments • Sound Healing Info • Healing Sounds Blog • Sound Healing Discussion • Healing Sounds Releases • Articles • Video • Uncategorized • Articles and Interviews • Healing Sounds Videos • Vibratory Research • Healing Sounds Radio Shows • Sound Healers Association • World Sound Healing Day • Bibliography • Links • Events and Training • The Power of Sound Healing Online Course • Tibetan Secrets of Happiness • Jonathan Goldman’s Complete Chakra Chanting Experience • The 11:11 Divine Name Seminar • Healing Sounds Correspondence Course • Workshops by Jonathan & Andi Goldman • Current Events • Contact Speaking of the resonance of our beloved planet, Jonathan has created two different sets of tuning forks which utilize the Schumann Resonance. This resonance, which was discovered and then validated over 60 years ago, is based upon the modulating frequency of the ionosphere of our planet. It is considered the frequency of the electro-magnetic field of the planet. While it modulates, depending upon conditions such as lightning and sun spots, it’s main frequency seems to be 7.83 Hz. Utilizin...

1) find the wavelength of the sound of the tuning fork (cm) 2) what's the vertical distance x from the top of the pipe to the antinode above the pipe (cm) 3) find the frequency of the tuning fork (Hz) Homework Equations velocity = wavelength . frequency for pipe closed at one end n = 1,3,5 wavelength for n times harmonic = 4.lenght/n The Attempt at a Solution I don't know for which harmonic the tuning fork exciting at 16.7 cm and 50.7 cm I assume for 16.7 is 1st harmonic so the wavelength = 4*16.7 = 66.8 cm since the 50.7 approximately 3 times 16.7, I assume that the n value is 3, which is the 2nd harmonic so the wavelength = 4*50.7/3 = 67.6 cm so the wavelength is 67 cm?? 2) can you give me clue, where to start? 3) f = v / wavelength f = 340 / 0.67 m = 507 Hz You seem to have done the hard parts OK. For (2) I think you only need to know the meaning of 'antinode'. Fix stuff in your head - don't leave this problem without sketching a picture of the amplitude against height on top of these pictures, which would also help answer. I guess I answered this question because I did that same experiment in English 'A' Level exam many years ago mnyah mnyah. (I got it a bit wrong because I forgot the /4 , but I did write my answer was unreasonable so probably got nearly full marks.) You seem to have done the hard parts OK. For (2) I think you only need to know the meaning of 'antinode'. Fix stuff in your head - don't leave this problem without sketching a picture of the amplitude agai...

If you have two tuning forks and you know that one tuning fork is 305 Hz, and when the tuning forks are pinged together you get a beat frequency of 6 Hz, then you know that the other tuning fork is either 311 Hz or 299 Hz. This idea I am fine with. Now, the second part is as follows: if you replace the known tuning fork with one that emits a higher frequency, the beat frequency is 4 Hz. What is the frequency of the second tuning fork? The answer is 311 Hz. But I don't understand how we can arrive there. The two things we know are $1.$ that the new frequency is $>$305 Hz and $2.$ that the beat frequency decreases from 6 Hz to 4 Hz. How does this then tie together that the other frequency is definitely 311 Hz? The way it was explained was that there are two situations: if the other frequency was 306 Hz (why? where did 306 Hz come from?) then the beat frequency would increase and if the other frequency was 311 Hz (again why?) then the beat frequency would decrease. Since we know the beat frequency decreases from 6 Hz to 4 Hz, it must be the second case. Well, you start with tuning fork A at 305 Hz and you know that B is either 299 Hz or 311 Hz. Now you take C, which is tuned at >305 Hz, which means at least 306 Hz (in integer steps...). If B was 299 Hz, then, together with C, which is higher tuned than A, the Beat frequency would increase to at least 306 Hz - 299 Hz = 7 Hz. But the beat frequency decreases, therefore, B must be 311 Hz. Thanks for contributing an answer to Phy...

This problem has been solved! You'll get a detailed solution from a subject matter expert that helps you learn core concepts. See Answer See Answer See Answer done loading Question:1.80 m is 0.500 kg A string is connected to a tuning fork whose frequency is 80.0 Hz and is held under tension by 0.500 kg. The tuning fork causes the string to vibrate as shown. The mass per unit length for the string is 3.85 . 10-2 kg/m. O 6.00 10 kg/m. O 6,80 10-3kg/m. 4.34 kg/m O 9.45 10-4 kg/m. 1.80 m is 0.500 kg A string is connected to a tuning fork whose frequency is 80.0 Hz and is held under tension by 0.500 kg. The tuning fork causes the string to vibrate as shown. The mass per unit length for the string is 3.85 . 10-2 kg/m. O 6.00 10 kg/m. O 6,80 10-3kg/m. 4.34 kg/m O 9.45 10-4 kg/m. Previous question Next question