An object in simple harmonic motion has an amplitude of 0.360 m and an oscillation period of 0.620 s. determine the maximum speed of the motion.

3.648 m/s

Explanation:

Data provided in the question:

Amplitude, A = 0.360 m

oscillation period, T = 0.620 s

The maximum speed of the motion, V = A × ω

where,

ω = Angular speed

also,

ω = (2π) / T

on substituting the respective values, we get

ω = (2π) / 0.620

or

ω = 10.134 rad / s

therefore,

V = 0.360 × 10.134 = 3.648 m/s

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s

false br