An object is undergoing shm with period 0.300 s and amplitude 6.00 cm. at t = 0 the object is instantaneously at rest at x = 6.00 cm. calculate the time it takes the object to go from x = 6.00 cm to x = -1.50 cm

Time taken is 0.087 s

Solution:

As per the question:

Time period, T = 0.300 s

Amplitude, A = 6.00 cm

Now,

To calculate the time taken:

For SHM, we know that:

                              (1)

At x = 6.00 cm, the object comes to rest instantaneously at times t = 0.00 s

Thus from eqn (1), for x = 6.00  cm:

Thus at t = 0.00 s, x = 6.00 cm

Now,

Using eqn (1) for x = - 1.50 cm:

We know that:

Thus

Time taken by the object in moving from x = 6.00 cm to x = 1.50 cm:

t' - t = 0.087 - 0.00 = 0.087 s

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer