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3 years ago

15

In certain ranges of a piano keyboard, more than one string is tuned to the same note to provide extra loudness. For example, th

e note at 110 Hz has two strings at this frequency. If one string slips from its normal tension of 596 N to 538.00 N, what beat frequency is heard when the hammer strikes the two strings simultaneously

1 answer:

pav-90 [236]3 years ago

3 0

Answer:

5.4893 beats/sec

Explanation:

We have f_1 and f_2 as frequency of two waves

also we know that according to question

$f_2=f_1\sqrt{\frac{F'}{F} }$

F and F' are forces on the two frequencies.

therefore, beat frequency is

$f=f_1-f_2=f_1-f_1\sqrt{\frac{F'}{F} }\\=110(1-\sqrt{\frac{538}{596} })\\=5.4893 \text{ beats/sec}$

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Point P is on the rim of a wheel of radius 2.0 m. At time t = 0, the wheel is at rest, and P is on the x-axis. The wheel undergo

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3 years ago

What an object is placed 8 mm from a concave spherical mirror a clear image can be projected on the screen 16 mm in front of me

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Mathematically, it can be written as

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(ii) In terms of the object's and image's distances.

The magnification of the spherical mirror is defined as the negative ratio of the image distance' $d_{i}$ ' to the object distance ' $d_{o}$ '.

Mathematically, it can be written as

$m= - \frac{d_{i}}{d_{o}} ------------(2)$

Now, from equation (1) and (2) we have,

$m = \frac{h_{i}}{h_{o}} = - \frac{d_{i}}{d_{o}} -----------(3)$

Given: Spherical Concave Mirror,

We will consider positive sign for object's and image's distance because both are in front of the mirror.

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Object's height $(h_{o}) = + 4 mm$

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