Physics: 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

THE SI TEMPERATURE SCALE IS KELVIN...

An object is undergoing shm with period 0.300 s - id03004668713, brandicarney70p8jlsp

kay4173

15.01.2022

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:

$x = Acos\omega t$ (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:

$6.00 = 6.00cos\omega t$

$cos\omega t = 1$

$\omega t = cos^{- 1}(1)$

$\omega t = 0$

Thus at t = 0.00 s, x = 6.00 cm

Now,

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

$- 1.50 = 6.00cos\omega t'$

$cos\omega t' = -0.25$

We know that:

$\omega = \frac{2\pi}{T}$

Thus

$\frac{2\pi}{0.300} t' = cos^{- 1}(0.25)$

$t' = 0.087\ s$

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

0,0(0 оценок)