Question

A mass is placed on a frictionless, horizontal table. A spring ( k = 185 k=185 N/m), which can be stretched or compressed, is placed on the table. A 4.5-kg mass is anchored to the wall. The equilibrium position is marked at zero. A student moves the mass out to x = 7.0 x=7.0 cm and releases it from rest. The mass oscillates in simple harmonic motion. Find the position, velocity, and acceleration of the mass at time t = 3.00 t=3.00 s. Round all of your answers to one decimal place.

Answer 1

The angular frequency of a mass on spring which will be given by -

= k / m

where, k = spring constant = 185 N/m

m = mass of an object = 4.5 kg

then, we get

= [(185 N/m) / (4.5 kg)]

= 41.1 rad²/s²

= 6.41 rad/s

For a simple harmonic motion (SHM), we have

x (t) = A cos t

(i) The position of a mass at time, t = 3 sec which will be given as -

x = (7 cm) cos [(6.41 rad/s) (3 s)]

x = (7 cm) cos (19.23 rad)

x = [(7 cm) (0.9442 rad)]

x = 6.6094 cm

x 6.61 cm

(ii) The velocity of a mass at time, t = 3 sec which will be given as -

x (t) = A cos t

Differentiating an above eq. w.r.t time & we get

(dx / dt) = d[A cos t] / dt

v = - A sin t

v = - [(6.41 rad/s) (7 cm)] sin [(6.41 rad/s) (3 s)]

v = - (44.87 cm/s) sin (19.23 rad)

v = - [(44.87 cm/s) (0.3293)]

v = - 14.7 cm/s

(iii) The acceleration of a mass at time, t = 3 sec which will be given as -

v = - A sin t

Differentiating an above eq. w.r.t time & we get

(dv / dt) = d[- A sin t] / dt

a = - ² A cos t

a = - [(6.41 rad/s)² (7 cm)] cos [(6.41 rad/s) (3 s)]

a = - (41.0881 cm/s²) cos (19.23 rad)

a = - [(41.0881 cm/s²) (0.9442 rad)]

a = - 38.7 cm/s²