Amass of 0.56 kg is attached to a spring and set into oscillation on a horizontal frictionless surface. the simple harmonic motion of the mass is described by x(t) = (0.42 m)cos[(6 rad/s)t]. determine the following. (a) amplitude of oscillation for the oscillating mass m (b) force constant for the spring n/m (c) position of the mass after it has been oscillating for one half a period m (d) position of the mass one-third of a period after it has been released m (e) time it takes the mass to get to the position x = −0.10 m after it has been released

(a) 0.42 m

(b) 20.16 N/m

(c) - 0.42 m

(d) - 0.21 m

(e) 17.3 s

Solution:

As per the question:

Mass, m = 0.56 kg

x(t) = (0.42 m)cos[cos(6 rad/s)t]

Now,

The general eqn is:

where

A = Amplitude

= angular frequency

t = time

Now, on comparing the given eqn with the general eqn:

(a) The amplitude of oscillation:

A = 0.42 m

(b) Spring constant k is given by:

Thus

(c) Position after one half period:

(d) After one third of the period:

(e) Time taken to get at x = - 0.10 m:

t = 17.3 s

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s

the temperature in kelvin of 2.0 dm3 of an ideal gas is doubled and its pressure is increased by a factor of four. what is the final volume of the gas