A 100 g object connected to a spring (k= 40 N/m) oscillates ✓ Solved

1. A 100 g object connected to a spring (k= 40 N/m) oscillates on a horizontal frictionless surface with an amplitude of 4.00 cm. Find the period and the total energy of the system.

2. What is the period of a pendulum with a length of 2 meters on Earth? On the Moon?

3. A 5 kg mass is attached to a spring that is hanging vertically. The spring is stretched 0.25 m from its equilibrium position. What is the spring constant?

4. A spring of constant k = 100 N/m hangs at its natural length from a fixed stand. A mass of 3 kg is hung on the end of the spring, and slowly let down until the spring and mass hang at their new equilibrium position. (a) Find the value of the quantity x in the figure above. The spring is now pulled down an additional distance x and released from rest. (b) What is the potential energy in the spring at this distance? (c) What is the speed of the mass as it passes the equilibrium position? (d) How high above the point of release will the mass rise? (e) What is the period of oscillation for the mass?

5. Calculate the length of a pendulum on earth whose frequency of oscillation is 10 Hz.

6. On planet X64J1, the period of a 0.50 m pendulum is 1.8 s. What is the acceleration due to gravity on this planet?

7. What is the value of g for a location where a pendulum 1.88 m long has a period of 2.20 s?

8. The period of a mass on a spring is 2 s. If k=50 N/m, what is the mass?

9. Explain SHM.

10. Explain Hook’s law.

Paper For Above Instructions

Simple Harmonic Motion (SHM) is a fundamental concept in physics that describes the oscillatory motion of an object in a restoring force proportional to its displacement from an equilibrium position. This type of motion is characterized by its periodic nature and can be exemplified by a variety of physical systems, such as springs and pendulums.

1. Period and Total Energy of a Spring-Mass System

For the 100 g object attached to a spring with a spring constant of k = 40 N/m, we can first determine the period (T) of the oscillation using the formula:

T = 2π√(m/k)

Here, m must be converted to kilograms: m = 100 g = 0.1 kg.

Substituting the values into the equation:

T = 2π√(0.1 kg / 40 N/m) = 2π√(0.0025) = 2π(0.05) ≈ 0.314 s

Next, we calculate the total mechanical energy of the system, which is given by the formula:

E = (1/2)kA²

Where A = 4.00 cm = 0.04 m.

Thus, substituting the values:

E = (1/2)(40 N/m)(0.04 m)² = (20)(0.0016) = 0.032 J

2. Period of a Pendulum

The period of a pendulum is given by:

T = 2π√(L/g)

For a pendulum of length L = 2 m on Earth (g ≈ 9.81 m/s²):

T_earth = 2π√(2/9.81) ≈ 2.84 s

On the Moon (g ≈ 1.62 m/s²):

T_moon = 2π√(2/1.62) ≈ 3.14 s

3. Spring Constant of Hanging Mass

For the 5 kg mass that stretches the spring by 0.25 m, we apply Hooke's Law, which states:

F = kx,

where F = mg.

Thus, setting k = F/x:

k = (5 kg)(9.81 m/s²) / 0.25 m ≈ 196.2 N/m

4. Mass and Spring System Analysis

For the spring with k = 100 N/m and a mass of 3 kg, we first find the displacement x:

At equilibrium, the weight of the mass equals the spring force:

mg = kx ⟹ x = mg/k = (3 kg)(9.81 m/s²) / 100 N/m ≈ 0.2943 m.

Next, to find the potential energy (PE) at an additional displacement x:

PE = (1/2)kx² = (1/2)(100)(0.2943)² ≈ 4.34 J.

For speed v at the equilibrium position:

Using conservation of energy, the total energy at maximum displacement equals the kinetic energy at equilibrium:

(1/2)kx² = (1/2)mv² ⟹ v = √(kx²/m) = √[(100)(0.2943)² / 3] ≈ 2.35 m/s.

To find the height above the release point:

Using energy conservation again, we set potential energy equal to gravitational potential energy:

PE = mgh ⟹ h = PE/(mg) = 4.34/(3*9.81) ≈ 0.147 m.

The period of oscillation is given by:

T = 2π√(m/k) = 2π√(3/100) ≈ 1.09 s.

5. Pendulum Length for 10 Hz Frequency

The frequency f of a pendulum is given by:

f = 1/T = 1/(2π√(L/g)) ⟹ L = g/(4π²f²).

Substituting g ≈ 9.81 m/s² and f = 10 Hz:

L = (9.81)/(4π²(10)²) ≈ 0.248 m.

6. Acceleration due to Gravity on Planet X64J1

Using the formula for a pendulum:

T = 2π√(L/g) ⟹ g = 4π²L/T².

Substituting L = 0.50 m and T = 1.8 s:

g = 4π²(0.50)/(1.8)² ≈ 9.25 m/s².

7. Value of g Based on Pendulum Period

For a pendulum of length 1.88 m with T = 2.20 s:

g = 4π²(1.88)/(2.20)² ≈ 9.82 m/s².

8. Mass of a Spring-Mass System

Using the period formula for a mass-spring system:

T = 2π√(m/k) ⟹ m = k(T/(2π))².

Substituting k = 50 N/m and T = 2 s:

m = 50(2/(2π))² ≈ 1.01 kg.

9. Explanation of Simple Harmonic Motion

Simple Harmonic Motion (SHM) is defined as a repetitive movement back and forth through an equilibrium position. It is periodic and characterized by constant amplitude, frequency, and period. Forces acting on the system are restoring forces that are proportional to the displacement, following Hooke’s Law.

10. Explanation of Hook's Law

Hook’s Law states that the force exerted by a spring is directly proportional to the distance it is stretched or compressed from its original length, mathematically expressed as F = kx, where k is the spring constant, and x is the displacement.

Conclusion

This paper has addressed various questions and calculations regarding simple harmonic motion, pendulums, and spring systems. Understanding these concepts is fundamental in the study of mechanics and oscillatory motion in physics.

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