Question

A 100 gm block is connected to a horizontal massless spring of force constant 25.6 N/m and is free to oscillate on a horizontal frictionless surface. The block is displaced on a horizontal frictionless surface. The block is displaced by 3 cm along positive x - direction and released with an initial velocity of $v_i=-16\sqrt{3}$ cm.

Express position of block as function of time.

• A. $x=3cos(16t+\frac{\pi }{2})cm$
• B. $x=3cos(16t+\frac{\pi }{3})cm$
• C. $x=3.5cos(16t+\frac{\pi }{6})cm$
• D. $x=3.2cos(16t+\frac{\pi }{4})cm$

Answer 1

The correct option is B $x=3cos(16t+\frac{\pi }{3})cm$

So far tile is at $x=3$ cm

And has $v=-16\sqrt{3}$ cm/s

Hmmmm....... So where will the particle be if its seen as a projection of a particle doing uniform

Circular motion.

Since the particle is coming back it will be at B

Keeping this in mind Let's start.

Given $m=0.1$ kg

$K=25.6$ N/m

We know $k=m{\omega }^2\Rightarrow \omega =16$

Particle is at $x=3$

$v=-16\sqrt{3}$

From the above diagram we know $\psi$ is greater then ${90}^{\circ }$