A vertical spring stretches 10 cm when a 5.0-kg block is suspended from its end.

Question

A vertical spring stretches 10 cm when a 5.0-kg block is suspended from its end. The
block is then displaced an additional 5.0 cm downward and released from rest to execute Simple Harmonic Motion. The block position as a function of time is given by:
(Take equilibrium position of spring-block system as origin and the upward-vertical
direction to be positive).

A) y = - 0.05 cos (9.9 t) m
B) y = - 0.15 sin (9.9 t ) m
C) y = - 0.10 cos (9.9 t ) m
D) y = - 0.10 sin (9.9 t + 9) m
E) y = - 0.15 sin (9.9 t + 5) m

Explanation / Answer

Since the spring is stretched for 5cm at equilibrium we have Amp = 5cm

Also at 10 cm stretch the spring balances 10 kg mass
kx =mg => k = 49/.1 = 490N/m

= (k/m) = 98 = 9.9

Now at t= 0 the mass is at it's negative extreme since upward direction is poistive

At T= 0 Extreme means cos

At T= 0 and negative extreme means -ve of cos = -cos

now we know every thing we get y= -0.50cos(9.9t)

A

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