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julia-pushkina [17]

4 years ago

14

A trombone has a variable length. When a musician blows into the mouthpiece and causes air in the tube of the horn to vibrate, t

he waves set up by the vibrations reflect back and forth in the horn to create standing waves. As the length of horn is made shorter, what happens to the frequency?
A) The frequency will increase or decrease depending on how hard the horn player blows.
B) The frequency will increase
C) The frequency will decrease
D) The frequency will increase or decrease depending on the diameter of the horn.
E) The frequency will remain same

1 answer:

tiny-mole [99]4 years ago

6 0

Answer:

B) The frequency will increase

Explanation:

The length of the instrument is given by

$L=\dfrac{\lambda}{2}\\\Rightarrow \lambda=2L$

Wavelength is given by

$\lambda=\dfrac{v}{f}$

$\dfrac{v}{f}=2L\ or\ L\propto \dfrac{1}{f}$

It can be seen that the length is inversely proportional to the frequency.

Here, the length of the trombone is decreasing so the frequency will increase.

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The density of the substance is<u> 10.5 g/cm³.</u>

The jewelry is made out of <u>Silver.</u>

Density ρ is defined as the ratio of mass <em>m</em> of the substance to its volume V<em>. </em> The cylinder contains a volume <em>V₁ of water</em> and when the jewelry is immersed in it, the total volume of water and the jewelry is found to be V₂.

The volume <em>V</em> of the jewelry is given by,

$V=V_2 -V_1$

Substitute 48.6 ml for <em>V₁ </em>and 61.2 ml for V₂.

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calculate the density ρ of the jewelry using the expression,

$\rho =\frac{m}{V}$

Substitute 132.6 g for <em>m</em> and 12.6 ml for <em>V</em>.

$\rho =\frac{m}{V}\\ =\frac{132.6 g}{12.6 ml} \\ =10.5 g/ml$

Since $1 ml=1 cm^3$ ,

The density of the jewelry is <u> 10.5 g/cm³.</u>

From standard tables, it can be seen that the substance used to make the jewelry is <u>silver</u><em><u>, </u></em>which has a density 10.5 g/cm³.

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How does a sound wave transfer energy to your ears ?

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3 years ago

a particle is moving with shm of period 8.0s and amplitude 5.0cm. find (a) the speed of particle when it is 3.0m from the centre

Fudgin [204]

Answer:

a) $speed=\pi cm/s$

b) $v_{max}=\frac{5\pi}{4} cm/s$

c) $a_{max}=\frac{5\pi^{2}}{16} cm/s^{2}$

Explanation:

The very first thing we must do in order to solve this problem is to find an equation for the simple harmonic motion of the given particle. Simple harmonic motion can be modeled with the following formula:

$y=Asin(\omega t)$

where:

A=amplitude

$\omega$ = angular frequency

t=time

we know the amplitude is:

A=5.0cm

and the angular frequency can be found by using the following formula:

$\omega=\frac{2\pi}{T}$

so our angular frequency is:

$\omega=\frac{2\pi}{8s}$

$\omega=\frac{\pi}{4}$

so now we can build our equation:

$y=5sin(\frac{\pi}{4} t)$

we need to find the speed of the particle when it is 3m from the centre of its motion, so we need to find the time t when this will happen. We can use the equation we just found to get this value:

$y=5sin(\frac{\pi}{4} t)$

$3=5sin(\frac{\pi}{4} t)$

so we solve for t:

$sin(\frac{\pi}{4} t)=\frac{3}{5}$

$\frac{\pi}{4} t=sin^{-1}(\frac{3}{5})$

$t=\frac{4}{\pi}sin^{-1}(\frac{3}{5})$

you can directly use this expression as the time or its decimal representation:

t=0.81933

since we need to find the speed of the particle at that time, we will need to get the derivative of the equation that represents the particle's position, so we get:

$y=5sin(\frac{\pi}{4} t)$

$y'=5cos(\frac{\pi}{4} t)*\frac{\pi}{4}$

which simplifies to:

$y' =\frac{5\pi}{4}cos(\frac{\pi}{4} t)$

and we can now substitute the t-value we found previously, so we get:

$y'=\frac{5\pi}{4}cos(\frac{\pi}{4} (0.81933))$

$y'=\pi$

so its velocity at that point is $\pi$ cm/s

b) In order to find the maximum velocity we just need to take a look at the velocity equation we just found:

$y' =\frac{5\pi}{4}cos(\frac{\pi}{4} t)$

its amplitude will always give us the maximum velocity of the particle, so in this case the amplitude is:

$A=\frac{5\pi}{4}$

so:

$v_{max}=\frac{5\pi}{4} cm/s$

c) we can use a similar procedure to find the maximum acceleration of the particle, we just need to find the derivative of the velocity equation and determine its amplitude. So we get:

$y'= \frac{5\pi}{4}cos(\frac{\pi}{4} t)$

We can use the chain rule again to find this derivative so we get:

$y" =-\frac{5\pi}{4}sin(\frac{\pi}{4} t)*(\frac{pi}{4})$

so when simplified we get:

$y"=-\frac{5\pi^{2}}{16}sin(\frac{\pi}{4} t)$

its amplitude is:

$A=\frac{5\pi^{2}}{16}$

so its maximum acceleration is:

$a_{max}=\frac{5\pi^{2}}{16} cm/s^{2}$

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