Question

(question1- )A 5 kg mass is attached to a spring on a horizontal frictionless surface. the elastic constant of the spring is 48.7 n/m if the mass is 31.6 cm right (+) of the equilbrium point and moving at speed 4.8 m/s find the total mechanical energy.? ( question2-) A 5 kg mass is attached to a spring on a horizontal frictionless surface. the elastic constant of the spring is 30.3 n/m .if the mass is 24.5 cm right (+) of the equilbrium point and moving at speed 6.7 m/s . what is the maximum distance from equilibrium( the amplitude in cm).? (question3- )a 5kg mass is attached to a spring on a horizontal frictionless surface. the elastic constant of the spring is 30.8 n/m if the mass is 28.5 cm right (+) of the equilbrium point and moving right (+)at speed 6.1 m/s what is the initial phase of the oscillation? rad ( question 4- ) ​a 10 kg mass is attached to a 250 n/m spring and set into vertical osillation. when the mass is 0.68 m above equilbrium it is moving at speed 3.4 m/s. find the amplitude of the osillation in (m)? can you please write the whole number for the answer because the computer will not accept short answer for example if the answer is 0.897654265 write like this please not like this 0.8 thanks a lot

Answer 1

Assuming the spring obeys Hooke's Law, then the mass will execute simple harmonic motion about the equilibrium position of the spring (see source).

The angular frequency of the oscillation is given by

w = sqrt(k/m)

where k is the spring constant, and m is the mass of object subjected to the force exerted by the spring. In this case,

w = sqrt(6 N/m / 0.344 kg) = 4.176 sec^-1

The period is related to the angular frequency by:

Period = 2*pi/w

so in this case,

Period = 2*pi/sqrt(6 N/m / 0.344 kg) = 1.504 sec <- This is the answer to part (a).

The potential energy of a spring-mass system is given by:

U = 0.5*k*x^2

Initially, the mass is at rest so all the energy in the system is potential energy. We are told that initially, the mass is displaced by 8 cm from the equilibrium position, so

U = 0.5 * 6 N/m * (0.08 m)^2 = 0.019 J

The mass will have its maxumum speed when the potential energy of the mass is zero, and all the energy in the system is present as kinetic energy (= 0.5*m*v^2). (This occurs when the mass is in the equilibrium position of the spring, x = 0). At that point:

0.5 * 6 N/m * (0.08 m)^2 = 0.5 * 0.344 kg * (v_max)^2

(v_max)^2 = (6 N/m * (0.08 m)^2)/(0.344 kg)

(v_max) = 0.334 m/s <- This is the answer to part (b)

For part (c), we start with Hooke's Law:

F = -k*x

and combine it with Newton's second law:

F = m*a

to obtain:

m*a = -k*x

a = -(k/m)*x

Both k and m are constants, so the acceleration is directly proportional to the displacement of the mass from the equilibrium position of the spring. The maximum acceleration will occur at the maximum displacement, which in this problem is at +/-8 cm (presumably the question is asking about the maximum *magnitude* of the acceleration).

We have, then, that:

a_max = -(6 N/m / 0.344 kg) * (+/-0.08 m)

|a_max| = 1.395 m/sec^2