Question

. A block of mass 4.0 kg on a spring has displacement as a function of time given by, x (t) = (1.0 cm) cos [(2.0 rad/s) t + 0.25] Determine: (i) The position of the block at 0.5 s. (0.31 cm) (ii) The period of the motion. (3.14 s) (iii) The spring constant of the spring. (16.0 N/m) (iv) The maximum speed of the block. (0.02 m/s) (v) The maximum force on the block. (0.16 N)

Answer 1

Answer and Explanation:

1. Evaluate the function x(t) at t=0.5

$x(t)=1cos(2t+0.25)\\x(0.5)=cos(2*0.5+0.25)=cos(1+0.25)=cos(1.25)=0.3153223624\approx0.31cm$

2. The period of motion T can be calculated as:

$T=\frac{2\pi}{\omega}$

Where:

$\omega=2rad/s$

So:

$T=\frac{2\pi}{2}=\pi\approx3.14s$

3. The angular frequency can be expressed as:

$\omega=\sqrt{\frac{k}{m} }$

Solving for k:

$k=(\omega)^2*m=(2)^2*4=4*4=16\frac{N}{m}$

4. Find the derivate of x(t):

$\frac{dx}{dt} =v(t)=-2sin(2t+0.25)$

Now, the sine function reach its maximum value at π/2 so:

$2t+025=\frac{\pi}{2}$

Solving for t:

$t=\frac{\frac{\pi}{2} -0.25}{2} =0.6603981634s$

Evaluating v(t) for 0.6603981634:

$v(0.6603981634)=-2sin(2*0.6603981634+0.25)=-2sin(\frac{\pi}{2} )=-2*1=2$

So the maximum speed of the block is:

$v(0.6603981634)=2cm/s=0.02m/s$

In the negative direction of x-axis

5. The force is given by:

$F=kx$

The cosine function reach its maximum value at 2π so:

$2t+0.25=2\pi$

Solving for t:

$t=\frac{2\pi-0.25}{2} =3.016592654s$

Evaluating x(t) for 3.016592654:

$x(3.016592654)=cos(2*3.016592654+0.25)=cos(2\pi)=1cm=0.01m$

Therefore the the maximum force on the block is:

$F=16*0.01=0.16N$