Question

Please explain how they got those answers thank you!

5. A mass m (m = 0.10 kg) is attached to a spring with spring constant k = 200 N/m. It is initially at its equilibrium point. It is then displaced a distanoe position, then released. What is the speed a = 5 cm from the equilibrium when it passes the equilibrium position? (a) 5.29 m/s (b) 7.95 m/s (c) 0.32 m/s (d) 1.43 m/s (e) 2.23 m/s 6. A harmonic oscillator with spring constant k and mass m has a frequency w e oscillator is displaced x = a, then at time t = 0 is released, what is the correct formula that is mathematically defined to be w = ym. If th from x = 0 to for the subsequent position as a function of time, r(t) ? (a) x = a sin(wt) (b) x = a + tot +lat2 (c) x = a(cos2 (at)-sin2 (at)) (d) x = a exp(-w2t2) (e) x-a cos(wt) ← 7. The pressure increases by 1.0 × 104 Pa for every meter of depth beneath t

Answer 1

a) energy conservation

1/2 * 200 * ( 0.05)^2 = 1/2*.10 *v^2

v = 2.23 m/s

b) because at t= 0 your displacement  shouuld be = a and it shoud be a SHM says the question so  

option b and c and d are not SHM

because they are eqution in t^2  

so which left us two option a and d

but t = 0 option a gives y = 0 so  

your answer will be e