A 0.30-kg object connected to a light spring with a force constant of 22.6 N/m oscillates on a frictionless horizontal surface. The spring is compressed 4.0 cm and released from rest. (a) Determine the maximum speed of the object. m/s (b) Determine the speed of the object when the spring is compressed 1.5 cm. m/s (c) Determine the speed of the object as it passes the point 1.5 cm from the equilibrium position. m/s (d) For what value of x does the speed equal one-half the maximum speed

(a)  vmax = 0.34m/s

(b)  v = 0.13m/s

(c)  v = 0.31m/s

(d)  x = 0.039m

Explanation:

Given information about the spring-mass system:

m: mass of the object = 0.30kg

k: spring constant = 22.6 N/m

A: amplitude of the motion = 4.0cm = 0.04m

(a) The maximum speed of the object is given by the following formula:

      (1)

w: angular frequency of the motion.

The angular frequency is calculated with the following relation:

          (2)

You replace the expression (2) into the equation (1) and replace the values of the parameters:

The maximum speed of the object is 0.34 m/s

(b) If the object is compressed 1.5cm the amplitude of its motion is A = 0.015m, and the maximum speed is:

The speed is 0.13m/s

(c) To find the speed of the object when it passes the point x=1.5cm, you first take into account the equation of motion:

You solve the previous equation for t:

With this value of t, you can calculate the speed of the object with the following formula:

The speed of the object for x = 1.5cm is v = 0.31 m/s

(d) To calculate the values of x on which v is one-half the maximum speed, you first calculate the time t:

The position will be:

The position of the object on which its speed is one-half its maximum velocity is 0.039

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures