A spring with a 10-kg mass and a damping constant 15 can be held stretched 1 meters beyond its natural length by a force of 2 newtons. Suppose the spring is stretched 2 meters beyond its natural length and then released with zero velocity. In the notation of the text, what is the value c^2 - 4mk? m^2 kg^2/sec^2. Find the position of the mass, in meters, after t seconds. Your answer should be a function of the variable t of the form c_1 e^alpha t + c_2 e^beta t where alpha = (the larger of the two) beta = (the smaller of the two) c_1 = c_2 =.

7.5=−k(2.5)−7.52.5=kma=−7.5x2.5my′′+9y′+−3y=0,y(0)=2.5,y(5)=0−9±81+4(m)(3)−−−−−−−−−−−√2m−92m±81+12m−−−−−−−−√2my=Ae−(9/2)xcos(81+12m−−−−−−−−√2mx)+Be−(9/2)xsin(81+12m−−−−−−−−√2mx)2.5=A+B⋅00=(2.5)e−45/2cos(81+12m−−−−−−−−√52m)+Be−45/2sin(81+12m−−−−−−−−√52m)

Step-by-step explanation:

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the solution of the equation is 2.23 ⇒ answer b

step-by-step explanation:

* lets revise the meaning of exponential function

- the form of the exponential function is y = ab^x, where a ≠ 0, b > 0 ,  

  b ≠ 1, and x is any real number

- it has a constant base b

- it has a variable exponent x

- to solve an exponential equation, take the log or ln of both sides,  

  and solve for the variable

* lets solve the problem

∵

- the base is 7 and the exponent is x

- insert ㏑ in both sides

∴

- use the rule

∴ x ㏑(7) = ㏑(77)

- to find x divide both sides by ㏑(7)

∴

* the solution of the equation is 2.23 to the nearest 2 decimal places