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3 years ago

7

A guitar string with a linear density of 2.0 g/m is stretched between supports that are 60 cm apart. The string is observed to f

orm a standing wave with three antinodes when driven at a frequency of 420 Hz.
Part A
What is the frequency of the fifth harmonic of this string?

1 answer:

Butoxors [25]3 years ago

3 0

<h2>The frequency of driver is 700 Hz</h2>

Explanation:

The frequency of wave in a string is given by the relation

n = $\frac{p}{2l} \sqrt{\frac{T}{m} }$

here n is the frequency

p is the number of antinodes and l is the length of string .

T is the tension in string and m is the mass per unit length

Thus 420 = $\frac{3}{120} \sqrt{\frac{T}{2} }$ I

Now if there is 5 antinodes , the value of p = 5

Thus n = $\frac{5}{120} \sqrt{\frac{T}{2} }$ II

Dividing II by I , we have

n/420 = 5/3

or n = 5/3 x 420 = 700 Hz

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3 years ago

It is easier to drive In sharp nails in a piece of wood than the blunt ones why?

murzikaleks [220]

The sharp nail has a less surface area in comparison to a blunt nail and pressure is inversely proportional to area so it is easier to Hamer a sharp nail into a wood rather than having a blunt nail in wood

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3 years ago

which statements about velocity are true? check all that apply. a. for velocity, you must have a number, a unit, and a direction

satela [25.4K]

a). for velocity, you must have a number, a unit, and a direction.
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c). the symbol for velocity is .
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you make its meaning very clear, so that everybody knows
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3 0

3 years ago

Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m

Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

$A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}$ .....(I)

$tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}$ ....(II)

Put the value of $\omega=0.628\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}$

$A=0.0198$

From equation (II)

$tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}$

$d=0.0023$

Put the value of $\omega=3.14\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}$

$A=0.0203$

From equation (II)

$tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}$

$d=0.0120$

Put the value of $\omega=6.28\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}$

$A=0.0209$

From equation (II)

$tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}$

$d=0.0257$

Put the value of $\omega=9.42\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}$

$A=0.0217$

From equation (II)

$tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}$

$d=0.0413$

Hence, This is the required solution.

5 0

3 years ago

A beam of light strikes a sheet of glass at an angle of 56.6° with the normal in air. You observe that red light makes an angle

Yuri [45]

Answer:

(a). Index of refraction are $n_{red}$ = 1.344 & $n_{violet}$ = 1.406

(b). The velocity of red light in the glass $v_{red} =$ 2.23 × $10^{8} \ \frac{m}{s}$

The velocity of violet light in the glass $v_{violet} =$ 2.13 × $10^{8} \ \frac{m}{s}$

Explanation:

We know that

Law of reflection is

$n_1 \sin\theta_{1} = n_2 \sin\theta_{2}$

Here

$\theta_1$ = angle of incidence

$\theta_2$ = angle of refraction

(a). For red light

1 × $\sin 56.6$ = $n_{red}$ × $\sin 38.4$

$n_{red}$ = 1.344

For violet light

1 × $\sin 56.6$ = $n_{violet}$ × $\sin 36.4$

$n_{violet}$ = 1.406

(b). Index of refraction is given by

$n = \frac{c}{v}$

$n_{red}$ = 1.344

$v_{red} = \frac{c}{n_{red} }$

$v_{red} = \frac{3(10^{8} )}{1.344}$

$v_{red} =$ 2.23 × $10^{8} \ \frac{m}{s}$

This is the velocity of red light in the glass.

The velocity of violet light in the glass is given by

$v_{violet} = \frac{3(10^{8} )}{1.406}$

$v_{violet} =$ 2.13 × $10^{8} \ \frac{m}{s}$

This is the velocity of violet light in the glass.

8 0

3 years ago