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4 years ago

6

A stretched string is 2.11 m long and has a mass of 19.5 g. When the string oscillates at 440 Hz , which is the frequency of the

standard A pitch, transverse waves with a wavelength of 15.3 cm travel along the string. Calculate the tension T in the string.

1 answer:

blsea [12.9K]4 years ago

5 0

Answer:

The tension of the string is 41.876 N

Explanation:

Given;

length of the string, L = 2.11 m

mass of the string, m = 19.5 g = 0.0195 kg

frequency of the wave, f = 440 Hz

wavelength, λ = 15.3 cm = 0.153 m

The velocity of the wave is given by;

v = fλ

v = 440 x 0.153

v = 67.32 m/s

Also the velocity of the wave is given by

$v = \sqrt{\frac{T}{\mu} }$

where;

μ is mass per unit length = 0.0195 / 2.11 = 0.00924 kg/m

T is the tension of the string

T = v²μ

T = (67.32)²(0.00924)

T = 41.876 N

Therefore, the tension of the string is 41.876 N

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Explanation:

We have given

Mass of the earth, $M=6\times 10^{24}kg$

Mass of the satellite, m = 5600 kg

Radius of earth, $R=6.4\times 10^6m$

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We know that $\omega =\frac{2\pi }{T}=\frac{2\times 3.14}{6200}=0.00101rad/sec$

Now

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(b)

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3 years ago

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∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Data:

m=90kg : crate mass

F= 282 N

μk =0.351 :coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

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W= m*g

W= 90kg*9.8 m/s²

W= 882 N

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N : Normal force (N)

Problem development

We apply the formula (1)

∑Fy = m*ay , ay=0

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N = 882 N

We replace the data in the formula (2)

Ff= μk*N = 0.351* 882 N

Ff= 309.58 N

We apply the formula (1) in x direction:

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a= (282 N - 309.58 N ) / (90)

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Kinematics of the crate

Because the crate moves with uniformly accelerated movement we apply the following formula :

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Where:

d:displacement in meters (m)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s²

Data

v₀ = 0.850 m/s

d = 0.75 m

a= - 0.306 m/s²

We replace the data in the formula (3)

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Answer:

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4 years ago