(a) Let <em>x</em> be the maximum elongation of the spring. At this point, the glider would have zero velocity and thus zero kinetic energy. The total work <em>W</em> done by the spring on the glider to get it from the given point (4.00 cm from equilibrium) to <em>x</em> is
<em>W</em> = - (1/2 <em>kx</em> ² - 1/2 <em>k</em> (0.0400 m)²)
(note that <em>x</em> > 4.00 cm, and the restoring force of the spring opposes its elongation, so the total work is negative)
By the work-energy theorem, the total work is equal to the change in the glider's kinetic energy as it moves from 4.00 cm from equilibrium to <em>x</em>, so
<em>W</em> = ∆<em>K</em> = 0 - 1/2 <em>m</em> (0.835 m/s)²
Solve for <em>x</em> :
- (1/2 (160 N/m) <em>x</em> ² - 1/2 (160 N/m) (0.0400 m)²) = -1/2 (0.190 kg) (0.835 m/s)²
==> <em>x</em> ≈ 0.0493 m ≈ 4.93 cm
(b) The glider attains its maximum speed at the equilibrium point. The work done by the spring as it is stretched away from equilibrium to the 4.00 cm position is
<em>W</em> = - 1/2 <em>k</em> (0.0400 m)²
If <em>v</em> is the glider's maximum speed, then by the work-energy theorem,
<em>W</em> = ∆<em>K</em> = 1/2 <em>m</em> (0.835 m/s)² - 1/2 <em>mv</em> ²
Solve for <em>v</em> :
- 1/2 (160 N/m) (0.0400 m)² = 1/2 (0.190 kg) (0.835 m/s)² - 1/2 (0.190 kg) <em>v</em> ²
==> <em>v</em> ≈ 1.43 m/s
(c) The angular frequency of the glider's oscillation is
√(<em>k</em>/<em>m</em>) = √((160 N/m) / (0.190 kg)) ≈ 29.0 Hz
