All categories

3 years ago

Two piano strings produce beats. Frequencies of the strings are 1560 Hz. and 1563 Hz. Choose the correct frequency of the beats.

2 answers:

4 0

Well first of all, I have to point out that the correct frequency of the beats is not on the list of choices that you provided.

When sources with different frequencies combine, two beats are created. The beat frequencies are the sum and difference of the frequencies that produce them.

If the source frequencies are 1,560 Hz and 1,563 Hz, then the beat frequencies are

3 Hz and 3,123 Hz.

The 3 Hz is perceived as a slight 'vibrato' in the piano's tone, so that it sounds warmer and not like a flat sine-wave generator.

The 3,123 Hz doesn't propagate well through the structure of the piano, and isn't noticeable to anyone except an expert listener (like a piano tuner). Only the top 6 keys [out of 88] on a standard piano keyboard have frequencies of 3,123 Hz or higher.

kakasveta [241]3 years ago

3 0

Beat frequency is given by the difference of two frequencies played together

$f_{beat} = |f_1 - f_2|$

given that

$f_1 = 1560 Hz$

$f_2 = 1563 Hz$

Now

$f_{beat} = |1560- 1563| Hz$

$f_{beat} = 3 Hz$

You might be interested in

Use the work—energy theorem to solve each of these problems. You can use Newton's laws to check your answers. Neglect air resist

andreyandreev [35.5K]

Answer:

a) It is moving at $43.15\frac{m}{s^{2}}$ when reaches the ground.

b) It is moving at $101.44\frac{m}{s^{2}}$ when reaches the ground.

Explanation:

Work energy theorem states that the total work on a body is equal its change in kinetic energy, this is:

$W=K_f-K_i$ (1)

with W the total work, Ki the initial kinetic energy and Kf the final kinetic energy. Kinetic energy is defined as:

$K=\frac{mv^2}{2}$ (2)

with m the mass and v the velocity.

Using (2) on (1):

$W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}$ (3)

In both cases the total work while the objects are in the air is the work gravity field does on them. Work is force times the displacement, so in our case is weight (w=mg) of the object times displacement (d):

$W=Fd=wd=mgd$ (4)

Using (4) on (3):

$mgd=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}$ (5)

That's the equation we're going to use on a) and b).

a) Because the branch started form rest initial velocity (vi) is equal zero, using this and solving (5) for final velocity:

$v_f=\sqrt{\frac{2mgd}{m}}=\sqrt{2gd}=\sqrt{2*9.8*95}$

$v_f=43.15\frac{m}{s^{2}}$

b) In this case the final velocity of the boulder is instantly zero when it reaches its maximum height, another important thing to note is that in this case work is negative because weight is opposing boulder movement, so we should use -mgd:

$-mgd=-\frac{mv_i^2}{2}$

Solving for initial velocity (when the boulder left the volcano):

$v_i=\sqrt{\frac{2mgd}{m}}=\sqrt{2gd}=\sqrt{2*9.8*525}$

$v_i=101.44 \frac{m}{s^{2}}$

3 0

2 years ago

Ron has not used heroin for several days. He has a

katen-ka-za [31]

I wanna say withdrawal? Ron needs to go to rehab

6 0

2 years ago

A spring on Earth has a 0.500 kg mass suspended from one end and the mass is displaced by 0.3 m. What will the displacement of t

julia-pushkina [17]

To solve this problem we will apply the concepts related to the Force of gravity given by Newton's second law (which defines the weight of an object) and at the same time we will apply the Hooke relation that talks about the strength of a body in a system with spring.

The extension of the spring due to the weight of the object on Earth is 0.3m, then

$F_k = F_{W,E}$