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Radda [10]
3 years ago
15

A 1.2-kg mass suspended from a spring of spring constant 22 N.m-1 executes simple harmonic motion of amplitude 5 cm. What is the

period T and frequency f of this oscillation? A mass of 1.2 kg attached to a spring is in simple harmonic motion along the x-axis with a period T = 2.5 s. If the total energy of the system is 2.7 J, what is the amplitude of the oscillation? An 8.0-kg block is attached to a spring with a spring constant of . If the spring is stretched 3.0 cm from its equilibrium position and released from rest, the maxium velocity attained by the mass is? Enter question here...
Physics
1 answer:
iren2701 [21]3 years ago
6 0

Answer:

a)  T = 1,467 s , b)    A = 0.495 m , c)  v = 4.97 10⁻² m / s

Explanation:

The simple harmonic movement is described by the expression

        x = A cos (wt + Ф)

Where the angular velocity is

       w = √ k / m

a) Ask the period

Angular velocity, frequency and period are related

      w = 2π f = 2π / T

      T = 2π / w

      T = 2pi √ m / k

      T = 2π √ (1.2 / 22)

      T = 1,467 s

      f = 1 / T

      f = 0.68 Hz

b) ask the amplitude

The mechanical energy of a harmonic oscillator

        E = ½ k A²

       A = √2 E / k

       A = √ (2 2.7 / 22)

       A = 0.495 m

c) the mass changes to 8.0 kg

As released from rest Ф = 0, the equation remains

         x = A cos wt

        w = √ (22/8)

        w = 1,658

         x = 3.0 cos (1,658 t)

Speed ​​is

         v = dx / dt

         v = -A w sin wt

The speed is maximum when without wt = ±1

         v = Aw

         v = 0.03    1,658

         v = 4.97 10⁻² m / s

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To solve this problem we will apply the relationship between Newton's second law and Hooke's law, with which we will define the balance of the system. From the only unknown in that equation that will be the constant of the spring, we will proceed to find the period of vibration of the car.

We know from Hooke's law that the force in a spring is defined as

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Here k is the spring constant and x the displacement

While by Newton's second law we have that the Weight can be defined as

F = mg

Here m is the mass and g the gravity acceleration.

The total weight would be

F = (80+60+70)(9.8)

F = 2058 N

Each spring takes a quarter of the weight, then

F_s = \frac{2058}{4} = 514.5N

Since the system is in equilibrium the force produced by the weight in each spring must be equivalent to the force of the spring, that is to say

F_s = F

\frac{1}{4} mg = kx

514.5N = k(0.04)

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The total mass is equivalent as the sum of all the weights, then replacing we have that the Period is

T = 2\pi \sqrt{\frac{1000+80+60+70}{4( 12862.5)}}

T = 0.9635s

Therefore the period of vibration of the car as it comes to rest after the four get in is 0.9635s

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