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3 years ago

8

Consider the system consisting of two pendulums coupled by a spring. Write the equations of motion in matrix form. Use small ang

le approximation.

1 answer:

Flura [38]3 years ago

8 0

Answer:

hello your question is incomplete attached below is the complete question

and solution to the question

Explanation:

To write The equation of the motion in Matrix form

i)First we write the net torque equation for left mass

ii)Next we write the net torque equation for right mass

iii) represent the equation of motion in Matrix form

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Two point charges are fixed on the y axis a negative point charge q1-26 μC at y1 = +0.21 m and a positive point charge q2 art y2

11111nata11111 [884]

Answer:

$3.13\times 10^{-5} C$

Explanation:

We are given that

$q_1=26\mu C=26\times 10^{-6} C$

$1\mu C=10^{-6} C$

$y_1=0.21 m$

$q=8.1\mu C=8.1\times 10^{-6} C$

$y_2=0.39 m$

F=28 N

We have to find the magnitude of q2.

We know that

$F=\frac{kq_1q_2}{r^2}$

Where $k=9\times 10^9$

Using the formula

Force on charge q due to charge q1

$F_1=\frac{9\times 10^9\times 26\times 10^{-6}\times 8.1\times 10^{-6}}{(0.21)^2}$

$F_1=42.98 N$

Force on charge q due to point charge q2

$F_2=\frac{9\times 10^9\times q_2\times 8.1\times 10^{-6}}{(0.39)^2}$

$F_2=4.79\times 10^5 q_2$

$F=F_1-F_2$

$28=42.98-4.79\times 10^5q_2$

$4.79\times 10^5q_2=42.98-28=14.98$

$q_2=\frac{14.98}{4.79\times 10^5}=3.13\times 10^{-5} C$

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3 years ago

Where do daughter cells obtain their dna

garik1379 [7]

Answer:

The parent cell

Explanation:

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6 foot I hope I helped

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Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving w

gtnhenbr [62]

Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

$U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}$

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

x1 = 1.00 m

x = 5.00

$u(x) = \frac{-2}{x} + \frac{4}{x^2}$

At x = 1.00 m

$u(1) = \frac{-2}{1} + \frac{4}{1^2}$

= 4J

Kinetic energy = (1/2)mv²

$= \frac{1}{2} * 4(3)^2$

= 18J

Total energy will be =

4J + 18J = 22J

At x = 5

$u(5) = \frac{-2}{5} + \frac{4}{5^2}$

$= \frac{4-10}{25} = \frac{-6}{25} J$

= -0.24J

Kinetic energy =

$\frac{1}{2} * 4Vf^2$

= 2Vf²

Total energy =

2Vf² - 0.024

Using conservation of energy,

Initial total energy = final total energy

22 = 2Vf² - 0.24

Vf² = (22+0.24) / 2

$Vf = \sqrt{frac{22.4}{2}$

= 3.33 m/s

b) magnitude of force when x = 5.0m

$u(x) = \frac{-2}{x} + \frac{4}{x^2}$

$\frac{-du(x)}{dx} = \frac{-d}{dx} [\frac{-2}{x}+ \frac{4}{x^2}$

$= \frac{2}{x^2} - \frac{8}{x^3}$

At x = 5.0 m

$\frac{2}{5^2} - \frac{8}{5^3}$

$F = \frac{2}{25} - \frac{8}{125}$

= 0.016N

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It produces only virtual images is the answer

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