A mass weighing 32 pounds stretches a spring 2 feet. Determine the amplitude and period of motion if the mass is initially released from a point 1 foot above the equilibrium position with an upward velocity of 5 ft/s.

Amplitude = 0.7183m

Period of motion (T) = 1.566 seconds

Step-by-step explanation:

First of all, lets convert all the figures given in the question to standard units for ease of calculations.

1 pound = 0.454kg. Therefore, 32 pounds = 32 x 0.454kg = 14.528kg

Mass(M) = 14.528kg

Velocity (v) = 5ft/s; now, 1ft/s = 0.305m/s; so 5ft/s = 5x0.305m/s = 1.524m/s

So v=1.524m/s

Change in length of spring (^x) = 2ft. If 1ft = 0.3048, 2ft = 2x0.3048 = 0.6096m

Now,we know that Weight = Mg where g = 9.81m/s^2

Therefore, W = 14.528 x 9.81 = 142.52N

Now, ^F is the same as weight and

^F = k(^x) where k is the spring constant. Therefore k = ^F/^x

K= 142.52/0.6096 = 233.79N/m

Angular speed (ω) = (k/m) ^(1/2)

Therefore, ω = (233/14.528)^(1/2)

ω = 16.093^(1/2) = 4.012 rad/s

It is well known that ω = 2πf ; where f is frequency.

Therefore, f = ω/2π ; f= 4.012/2π = 0.6385 hz.

Period of motion (T) = 1/f

So, T = 1/0.6385 = 1.566 seconds

From proven oscillation equations;

V^2 = ω^2 (A^2 - x^2) ; where v is velocity, and A is amplitude.

Therefore let's make A the subject of the the equation ;

(V^2/ω^2) + (x^2) = (A^2)

So, {(1.524^2)/(4.012^2)} + (0.6096^2) = (A^2)

So, (A^2) = 0.5159

So, A = 0.5159^(1/2) = 0.7183m

answer: it is 1: 47 pm