Please help!! A 1.5-kg mass attached to an ideal massless spring with a spring c

Question

Please help!!

A 1.5-kg mass attached to an ideal massless spring with a spring constant of 20.0 N/m oscillates on a horizontal, frictionless track. At time t = 0.00 s, the mass is released from the rest at x = 10.0 cm. (That is, the spring is stretched by 10.0 cm.) Find the maximum amplitude A for this SHO. What is the phase angle? Find the frequency of the oscillations. Write the proper form of the displacement x as a function of time t(i.e. the equation of the postion of the SHO vs. time). From d, find an expression of the velocity of the SHO. From e, find an expression of the acceleration of the SHO. Determine the maximum speed of the mass. At what point in the motion dose the maximum speed occur? What is the maximum acceleration of the mass? At what point in the motion does the maximum acceleration occur? Determine the total energy of the oscillating system.

Explanation / Answer

The equation of motion is A sin(w*t + theta) = 10 sin(w*t +pi/2)

The maximum amplitude A for the SHM is 10 cm

The inital phase angle is pi/2 or 90 degrees.

cyclic frequency of oscillation = sqrt(k/m) = sqrt(20/1.5) =3.65 rad /sec

Frequency = cyclic frequency/(2*pi) =0.58 Hz

The velocity equation is the differential of the displacement equation = 10*w cos(w*t + pi/2) =36.5 * cos(3.65t+pi/2) cm/sec

The accelaration is the differentaion of velocity equation = -10 *w*w sin(w*t + pi/2) = -133.22 sin(3.65t+pi/2) cm/sec2

Maximum speed of the mass occurs at its mean positon when the spring is unstretched and is equal to 36.5 cm/sec

the max accelation of the mass occurs at the ends of the SHM motion, when it has the maximum displacement . This occurs at the two extreme ends and is equal to 133.22 cm/sec2

The total energy of the system at any time = initial Potential energy stored in the string = 0.5 k * x^2 = 0.5 * 20 *0.1 *0.1 =0.1J

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