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

Answer 2

a. The force (magnitude and direction) acting on the object 3.50 seconds after it is released is -11.0 Newton to the left.

b. The number of times the object oscillate in 3.50 seconds is 0.880.

Given the following data:

• Mass = 2.00 kg.

• Spring constant = 5.00 N/m.

• Displacement = 3.00 m.

• Time = 3.50 seconds.

a. To find the force (magnitude and direction) acting on the object 3.50 seconds after it is released:

Hooke's law force affects the motion of an object attached to the end of a horizontal spring on a frictionless horizontal track.

Mathematically, Hooke's law force is given by the formula:

$F = -kx$

Where:

• F is Hooke's law force.

• k is spring constant.

• x is the displacement.

Since the object initiate simple harmonic motion, we would determine its angular frequency (w):

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

Substituting the values into the above formula, we have:

$w = \sqrt{\frac{5}{2}}\\\\w = \sqrt{2.5}$

Angular frequency (w) = 1.58 rad/s

Next, we would determine the displacement of the object from its equilibrium position by using this formula:

$x = Acos(wt)\\\\x = 3cos(1.58[3.50])\\\\x = 3cos(5.53)\\\\x = 3(0.7295)$

x = 2.19 meters

From Hooke's law force:

$F = -5(2.19)$

Force, F = -10.95 ≈ -11.0 Newton

Therefore, the force is directed to the left (opposite direction) because it is negative.

b. To find how many times the object oscillate in 3.50 seconds:

Mathematically, the number of oscillation by an object is given by the formula:

$n = \frac{\Delta t}{2 \pi} \sqrt{\frac{k}{m}} \\\\n = \frac{3.5}{2 \pi} \sqrt{\frac{5}{2}}\\\\n = \frac{3.5}{2 \pi}(1.58)\\\\n = \frac{5.53}{2(3.142)}$

Number of oscillation, n = 0.880.

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