Question

PHYS 2048 In-Class Quiz 6: March 15th, 201 Name and N number: from a vertical spring. You pull the mass down from equilbrium. You give the A small mass is hanging at rest from a vertical spring. You pull the mass down from equilibrium. You give mass an initial velocity of magnitude vo, up. Call the maximum speed situation 2, you pull the mass to the same position, but give it an intitial velocity of the same magnitude w of the mass in this situation vi. For o, down. Call the maximum speed Compare the sizes of vo, vI and v2. Be sure to explain your ans of the mass in this situation v2 in 己150( 2 A mass is hanging at rest from a vertical spring. Use a coordinate system with the y axis pointing vertically up with y -0 at the equilibrium position. You pull the mass down from the equilibriun and give it an intitial velocity down. Label below which of the graphs correspond to the y component of the position, velocity and acceleration. Be certain to explain your answers. tine tise

Hi, I wanted to know if my first 3 questions were correct and the last two questions

Answer 1

1. ( remember all three quantities are speeds so they are positive numbers). In SHM, speed is maximum at the equilibrium position so $v_1$ and $v_2$ (max. speeds of the respective SHMs) must be greater than $v_0$ . Further, $v_1$ and $v_2$ are equal because the two situation belong to the SHM that are just out of phase.
     i) $y = - ~y_m ~Cos(\omega t + \phi)$
    ii) $y = -~y_m ~Cos(\omega t - \phi)$

(this can be proved using given initial conditions), where $\phi$ is phase constant to account for the fact that initial position in not extreme position because at extreme position velocity must be zero and magnitude of displacement maximum.

2. initial displacement is negative and its magnitude initially increase (i.e. displacement becomes more negative) because initial velocity is negative. Initial velocity negative but initial acceleration must be positive because in SHM direction of acceleration is opposite to direction of displacement (at all times). So magnitude of velocity initially decreases. Further since acceleration is proportion to -ve of the displacement, as magnitude of displacement increases (initially), the magnitude of acceleration increases. (Anyhow in spring motion force is opposite to displacement and larger the magnitude of displacement larger the magnitude of force, and hence that of the acceleration).

3. $k|x| = 80 \Rightarrow k = \frac{80}{(0.6 -0.1)} = 160~ Nm^{-1}$

   Amplitude = 0.5 m (released from rest, so initial extension is the maximum extension)

$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{k}{m_1 + m_2}} = \sqrt{\frac{160}{10}} = 4 ~rad~ s^{-1}$

4 (i) $x = 0.5 \hat{x}(m) ~Cos(4t); ~~~v_x = - 2\hat{x}(ms^{-1})~Sin(4t);~~~~~~a_x = - 8\hat{x}(ms^{-2})~Cos(4t)$

fert Colr) f-f = m, a anel Mia