Question

1A) A 0.3-kg block, attached to a spring, executes simple harmonic motion according to x = 0.08 cos (35 rad/s⋅t), where x is in meters and t is in seconds. Find the total energy of the spring-mass system.

Ans.E =1.18 J

1B) A 1.5-kg cart attached to an ideal spring with a force constant (spring constant) of 20 N/m oscillates on a horizontal, frictionless track. At time t = 0.00 s, the cart is released from rest at position x = 10 cm from the equilibrium point. Find the position of the cart and its velocity at t = 5.0 s

Ans. x=8.25 cm

Answer 1

1 A)

given
m = 0.3 kg
x = 0.08*cos(35*t)
compare the above equation with , x = A*cos(w*t)

amplitude, A = 0.08 m
angular frequency, w = 35 rad/s
maximum speed of the block, v_max = A*w

= 0.08*35

= 2.80 m/s

the total energy of the spring-mass system, U = (1/2)*m*v_max^2

= (1/2)*0.3*2.8^2

= 1.176 J

= 1.18 J (when rounded to three significant figures) <<<<<<------------Answer

1 B)

given
m = 1.5 kg
k = 20 N/m
A = x = 10 cm = 0.1 m

angular frequency of motion, w = sqrt(k/m)

= sqrt(20/1.5)

= 3.65 rad/s

we know, position of block,

x = A*cos(w*t)

at t = 5 s

x = 0.1*cos(3.65*5)

= 0.0825 m

= 8.25 cm <<<<<<<<------------------Answer

let v is the velocity of the block.

v = dx/dt

= A*(-sin(w*t))*w

= -A*w*sin(w*t)

at t = 5 s

v = -0.1*3.65*sin(3.65*5)

= 0.206 m/s (or) 20.6 cm/s <<<<<<<<------------------Answer

Note : use calculator in radian mode.