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emmainna [20.7K]
3 years ago
5

Q 10.17: Which of the following statements concerning simple harmonic motion is false? A : A restoring force acts on an object i

n simple harmonic motion that is directed in the same direction as the object's displacement. B : A restoring force acts on the object that is proportional to the object's displacement from its equilibrium position. C : During simple harmonic motion, the net force on the object is zero when it is at its equilibrium position. D : The displacement versus time graph for an object in simple harmonic motion resembles the sine or cosine function. E : The amplitude of the motion of an object in simple harmonic motion is the maximum distance the object moves from its equilibrium position.
Physics
1 answer:
Alenkasestr [34]3 years ago
7 0

Answer:

A : A restoring force acts on an object in simple harmonic motion that is directed in the same direction as the object's displacement.

Explanation:

Statement A is the false one:

A : A restoring force acts on an object in simple harmonic motion that is directed in the same direction as the object's displacement. --> FALSE. The restoring force in the simple harmonic motion is given by

F=-kx

where

k is the spring constant

x is the displacement of the system, measured with respect to the equilibrium position

As we can notice from the equation, there is a negative sign in front of (kx): this means that the force, F, and the displacement, x, have opposite directions. In fact, the restoring force of a simple harmonic oscillator always acts to restore the equilibrium position, therefore it acts in the opposite direction as that of the displacement.

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The barricade at the end of a subway line has a large spring designed to compress 2.00 m when stopping a 1.10 ✕ 105 kg train mov
Mrac [35]

Answer:

(a) k = 1684.38 N/m = 1.684 KN/m

(b) Vi = 0.105 m/s

(c) F = 1010.62 N = 1.01 KN

Explanation:

(a)

First, we find the deceleration of the car. For that purpose we use 3rd equation of motion:

2as = Vf² - Vi²

a = (Vf² - Vi²)/2s

where,

a = deceleration = ?

Vf = final velocity = 0 m/s (since, train finally stops)

Vi = Initial Velocity = 0.35 m/s

s = distance covered by train before stopping = 2 m

Therefore,

a = [(0 m/s)² - (0.35 m/s)²]/(2)(2 m)

a = 0.0306 m/s²

Now, we calculate the force applied on spring by train:

F = ma

F = (1.1 x 10⁵ kg)(0.0306 m/s²)

F = 3368.75 N

Now, for force constant, we use Hooke's Law:

F = kΔx

where,

k = Force Constant = ?

Δx = Compression = 2 m

Therefore.

3368.75 N = k(2 m)

k = (3368.75 N)/(2 m)

<u>k = 1684.38 N/m = 1.684 KN/m</u>

<u></u>

<u>(</u>c<u>)</u>

Applying Hooke's Law with:

Δx  = 0.6 m

F = (1684.38 N/m)(0.6 m)

<u>F = 1010.62 N = 1.01 KN</u>

<u></u>

(b)

Now, the acceleration required for this force is:

F = ma

1010.62 N = (1.1 kg)a

a = 1010.62 N/1.1 x 10⁵ kg

a = 0.0092 m/s²

Now, we find initial velocity of train by using 3rd equation of motion:

2as = Vf² - Vi²

a = (Vf² - Vi²)/2s

where,

a = deceleration = -0.0092 m/s² (negative sign due to deceleration)

Vf = final velocity = 0 m/s (since, train finally stops)

Vi = Initial Velocity = ?

s = distance covered by train before stopping = 0.6 m

Therefore,

-0.0092 m/s² = [(0 m/s)² - Vi²]/(2)(0.6 m)

Vi = √(0.0092 m/s²)(1.2 m)

<u>Vi = 0.105 m/s</u>

4 0
3 years ago
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