Question

A 0.50 kg air track glider is attached to the end of the track by a spring that has a constant of 20.0 N/m. The glider is compressed 15.0 cm from equilibrium and released at t = 0, so that it oscillates back and forth on the friction less air track surface. What is its magnitude of acceleration at a time equal to 1/8th of the oscillator's period? 

(A) 3 m/s² 

(B) 3√2 m/s² 

(C) 6 m/s² 

(D) 3√3 m/s²

Answer 1

Correct Answer is: (B) 3√2 m/s² 

Comments

How did the angle 45° came?

Answer 2

\(PR = \frac A{\sqrt 2}\)

Then \(\tan \theta = \frac LA\)

\(= \cfrac{\frac A {\sqrt2}}{\frac A{\sqrt 2}}\)

\(=1\)

\(\theta = 45°\)