Question

A 2.00-kg object on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 5.00 N/m. The object is displaced 3.00 m to the right from its equilibrium position and then released, initiating simple harmonic motion. (a) What is the force (magnitude and direction) acting on the object 3.50 s after it is released? (b) How many times does the object oscillate in 3.50 s?

Answer 1

Answer:

Force = 10.244 Newtons

b) No of oscillations = 0.88

Explanation:

Since the block executes SHM we can write it's position as function of time as

$x=Asin(\omega t)$

ω is the natural frequency of the system

A is the amplitude of the system

$\omega =\sqrt{\frac{k}{m}}$

Thus accleration of the block

$x=Asin(\omega t)\\\\a=\frac{d^{2}x(t)}{dt^{2}}\\\\a=\frac{d^{2}Asin(\omega t)}{dt^{2}}\\\\a=-A\omega ^{2}sin(\omega t)$

Thus using the given values at t= 3.50 sec we can calculate the acceleration as

$k=\sqrt{\frac{5}{2}}=1.58rad/sec\\\\A=3.0m\\\\a=-3.0\times (1.58)^{2}sin(1.58\times 3.5)\\\\a=5.122m/s^{2}$

thus force can be calculated using newtons second law as

$Force = m\times accleration\\\\force=2\times 5.122=10.244 Newtons$

b)

Now no of oscillations can be obtained as

$\frac{time}{TimePeriod}\\\\Time Period=\frac{2\pi }{w} \\\\time Period=\frac{2\pi }{1.58}\\\\ TimePeriod=3.976secs$

no of oscillations in 3.50 seconds = 3.50/3.976 = 0.88