Calculate the frequency of the tuning fork which when vibrated with the tuning fork of frequency \[256{\text{ }}Hz\] produces $6$ \[beat/s\] and when vibrated with tuning fork having frequency \[253{\text{ }}Hz\] produces \[3{\text{ }}beats/s\].
Answer
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Hint: Typically aluminum, contains a stem (handle) and two prongs that form a U-shaped fork The implement vibrates at a gaggle frequency after being struck on the heel of the hand and it’s wont to assess the vibratory sensation and hearing.
Complete step-by-step solution:
Let A and B be two tuning forks, then the frequency of A
\[nA{\text{ }} = {\text{ }}256{\text{ }}Hz\]
Tuning fork b produces beats with fork B.
\[nB{\text{ }} = {\text{ }}nA \pm n\]
\[ = {\text{ }}250{\text{ }}or{\text{ }}262{\text{ }}Hz\] \[ = {\text{ }}256 \pm 6\]
Fork B produces \[3{\text{ }}beat/s\] with the opposite fork (frequency\[253{\text{ }}Hz\]), it is only possible when. \[n3{\text{ }} = {\text{ }}250{\text{ }}Hz\]
Additional Information: Sound waves are produced by vibrating objects. Whether it's the sound of an individual's voice, the sound of a piano, the sound of a trombone, or the sound of a physics book slamming to the ground , the source of the sound is typically a vibrating object. An implement may be a useful illustration of how a vibrating object can produce sound. When the implement is hit a rubber hammer, and the tines begin to vibrate. The rear and forth vibration of the tunes produces disturbances of surrounding air molecules. As a tine stretches outward from its usual position, it compresses surrounding air molecules into alittle region of space; this creates a high-pressure region next to the tune. because the tune then moves inward from its usual position, the air surrounding the tune expands; this produces a low-pressure region next to the tune. The high-pressure regions are mentioned as compressions and thus the low-pressure regions are mentioned as rarefactions.
Note: As the tunes still vibrate, an alternating pattern of high and low-pressure regions is formed. These regions are transported through the encircling air, carrying the sound signal from one location to a special one.
Complete step-by-step solution:
Let A and B be two tuning forks, then the frequency of A
\[nA{\text{ }} = {\text{ }}256{\text{ }}Hz\]
Tuning fork b produces beats with fork B.
\[nB{\text{ }} = {\text{ }}nA \pm n\]
\[ = {\text{ }}250{\text{ }}or{\text{ }}262{\text{ }}Hz\] \[ = {\text{ }}256 \pm 6\]
Fork B produces \[3{\text{ }}beat/s\] with the opposite fork (frequency\[253{\text{ }}Hz\]), it is only possible when. \[n3{\text{ }} = {\text{ }}250{\text{ }}Hz\]
Additional Information: Sound waves are produced by vibrating objects. Whether it's the sound of an individual's voice, the sound of a piano, the sound of a trombone, or the sound of a physics book slamming to the ground , the source of the sound is typically a vibrating object. An implement may be a useful illustration of how a vibrating object can produce sound. When the implement is hit a rubber hammer, and the tines begin to vibrate. The rear and forth vibration of the tunes produces disturbances of surrounding air molecules. As a tine stretches outward from its usual position, it compresses surrounding air molecules into alittle region of space; this creates a high-pressure region next to the tune. because the tune then moves inward from its usual position, the air surrounding the tune expands; this produces a low-pressure region next to the tune. The high-pressure regions are mentioned as compressions and thus the low-pressure regions are mentioned as rarefactions.
Note: As the tunes still vibrate, an alternating pattern of high and low-pressure regions is formed. These regions are transported through the encircling air, carrying the sound signal from one location to a special one.
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