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3 years ago

For a spring oscillating in simple harmonic motion, at what point will the velocity of

Physics

1 answer:

Crazy boy [7]3 years ago

6 0

Answer:

equilibrium position.

Explanation:

In simple harmonic motion , velocity v(t) is given by,

v(t) = -ω A sin(ωt + φ)

where

ω = angular velocity of the corresponding circular motion

A = amplitude

t = time

φ = the initial angle of the corresponding circular motion when the motion begin.

v (t) get maximized when sin value is maximized , i.e. sin $\alpha$ =1

The particle has maximum speed when it passes through the equilibrium position.

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Consider a vibrating system described by the initial value problem. (A computer algebra system is recommended.) u'' + 1 4 u' + 2

GarryVolchara [31]

Answer:

Therefore the required solution is

$U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t$

Explanation:

Given vibrating system is

$u''+\frac{1}{4}u'+2u= 2cos \omega t$

Consider U(t) = A cosωt + B sinωt

Differentiating with respect to t

U'(t)= - A ω sinωt +B ω cos ωt

Again differentiating with respect to t

U''(t) = - A ω² cosωt -B ω² sin ωt

Putting this in given equation

$-A\omega^2cos\omega t-B\omega^2sin \omega t+ \frac{1}{4}(-A\omega sin \omega t+B\omega cos \omega t)+2Acos\omega t+2Bsin\omega t = 2cos\omega t$

$\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)cos \omega t+(-B\omega^2-\frac{1}{4}A\omega+2B)sin \omega t= 2cos \omega t$

Equating the coefficient of sinωt and cos ωt

$\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)= 2$

$\Rightarrow (2-\omega^2)A+\frac{1}{4}B\omega -2=0$ .........(1)

and

$\Rightarrow -B\omega^2-\frac{1}{4}A\omega+2B= 0$

$\Rightarrow -\frac{1}{4}A\omega+(2-\omega^2)B= 0$ ........(2)

Solving equation (1) and (2) by cross multiplication method

$\frac{A}{\frac{1}{4}\omega.0 -(-2)(2-\omega^2)}=\frac{B}{-\frac{1}{4}\omega.(-2)-0.(2-\omega^2)}=\frac{1}{(2-\omega^2)^2-(-\frac{1}{4}\omega)(\frac{1}{4}\omega)}$

$\Rightarrow \frac{A}{2(2-\omega^2)}=\frac{B}{\frac{1}{2}\omega}=\frac{1}{(2-\omega^2)^2+\frac{1}{16}\omega}$

$\therefore A=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega}$ and $B=\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega}$

Therefore the required solution is

$U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t$

5 0

3 years ago

A piece of putty and a tennis ball with the same mass are thrown against a wall with the same velocity. Which object experiences

cluponka [151]

Answer:

Explanation:

Firstly, we have to define momentum.

Momentum is define as the product of mass and velocity.

That is P = mass×velocity

Also considering the third law of motion which states that: For every action, there is equal and opposite reaction.

Moreso, considering the 2nd law of motion which states that the rate of change in the momentum of a body is equal to the applied force and takes place in the direction of the applied force.

Now, applying P = mass×velocity

They both have same mass and velocity definitely, they will both experience same momentum.

Also from the question, the both share same velocity hence, the will both hit the wall with same velocity meaning the will both feel the same impact from the wall as well. Hence the third law of motion proves this right.

5 0

2 years ago

Stability is a term that means...

Aleksandr-060686 [28]

The correct answer is A

7 0

3 years ago

What is the mass of a bicycle that has 90 kg.m/s of momentum at a speed of 6 m/s?

OLEGan [10]

Answer:

m = 15 kg

Explanation:

p = m × v

90 = m × 6

6m = 90

m = 90/6

m = 15 kg

6 0

2 years ago

Can you help me doing an essay about actual self

Rudik [331]

Answer:

yes absolutely

Explanation:

why not

4 0

2 years ago