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To calculate the extension of a spring with mass attached to it, you can use Hooke's Law, which states that the force exerted by the spring is directly proportional to the extension of the spring. The formula is F = kx, where F is the force applied, k is the spring constant, and x is the extension of the spring. By rearranging the formula, you can calculate the extension x = F / k.
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What is the relationship between extension and mass?
The relationship between extension and mass is described by Hooke's Law, which states that the extension of a spring is directly proportional to the force applied to it, as long as the elastic limit of the material is not exceeded. This means that the greater the mass attached to the spring, the more it will stretch. The relationship can be expressed mathematically as F = kx, where F is the force applied, k is the spring constant, and x is the extension of the spring.
How does mass affect the period of a spring?
The period of a spring is not affected by its mass. The period of a spring is determined by its stiffness and the force applied to it, not by the mass of the object attached to it.
A frequency of vibration of a mass-spring system is 5Hz when a 4g mass is attached to the spring .Calculate the forces constant of the spring?
To calculate the force constant of the spring (k), you can use the formula for the frequency of vibration of a mass-spring system: f = 1 / (2π) * √(k / m) where f is the frequency, k is the force constant of the spring, and m is the mass. Rearranging the formula gives: k = (4π^2 * m * f^2). Plugging in the given values: k = (4π^2 * 0.004 * 5^2) ≈ 1.256 N/m.
What factors influence the period of a spring?
The period of a spring is influenced by factors such as the mass attached to the spring, the spring constant, and the amplitude of the oscillation.
What is the effective mass of spring?
The effective mass of a spring is the mass that would behave the same way as the spring when subjected to a force or acceleration. It is a concept used in physics to simplify calculations in systems involving springs. The effective mass of a spring depends on its stiffness and the mass it is attached to.
Predict how many centimeters the spring will stretch if a total mass of 700 grams were attached?
To predict how many centimeters the spring will stretch, we need to know the spring constant in N/cm and apply Hooke's Law. Hooke's Law states that the force exerted by a spring is directly proportional to its extension. By knowing the spring constant and the total mass attached, we can calculate the stretch.
What is the relationship between extension and mass?
The relationship between extension and mass is described by Hooke's Law, which states that the extension of a spring is directly proportional to the force applied to it, as long as the elastic limit of the material is not exceeded. This means that the greater the mass attached to the spring, the more it will stretch. The relationship can be expressed mathematically as F = kx, where F is the force applied, k is the spring constant, and x is the extension of the spring.
Which states the relationship between the mass attached to the end of the spring and the length the spring is stretched?
more mass the longer the spring
How does mass affect the period of a spring?
The period of a spring is not affected by its mass. The period of a spring is determined by its stiffness and the force applied to it, not by the mass of the object attached to it.
In a seismometer the pen is attached to an inertial mass suspended by a spring What is the rotating drum attached to?
the ground
A frequency of vibration of a mass-spring system is 5Hz when a 4g mass is attached to the spring .Calculate the forces constant of the spring?
To calculate the force constant of the spring (k), you can use the formula for the frequency of vibration of a mass-spring system: f = 1 / (2π) * √(k / m) where f is the frequency, k is the force constant of the spring, and m is the mass. Rearranging the formula gives: k = (4π^2 * m * f^2). Plugging in the given values: k = (4π^2 * 0.004 * 5^2) ≈ 1.256 N/m.
What factors influence the period of a spring?
The period of a spring is influenced by factors such as the mass attached to the spring, the spring constant, and the amplitude of the oscillation.
What is the effective mass of spring?
The effective mass of a spring is the mass that would behave the same way as the spring when subjected to a force or acceleration. It is a concept used in physics to simplify calculations in systems involving springs. The effective mass of a spring depends on its stiffness and the mass it is attached to.
Why the mass of the spring had not necessarily been taken into consideration in the spring elastisity experiment?
In the spring elasticity experiment, the mass of the spring is often neglected because the mass of the spring itself is usually negligible compared to the masses being hung on it. Additionally, the focus of the experiment is typically on the relationship between the force applied to the spring and the resulting extension, rather than the mass of the spring.
How would k vary with different mass of trapped gas?
The value of the spring constant ''k'' in a spring-mass system would remain constant regardless of the mass of the trapped gas, as it only depends on the stiffness of the spring and not on the mass attached to it.
Is the period of a vibrating spring linearly affected by the mass of the attached object?
Let a mass m be attached to the end of a spring with spring constant k. The spring extends and comes to rest with an equilibrium extension e. At equilibrium; Weight = Force exerted by spring => mg = ke -------- 1 Suppose the spring is displaced through a displacement x downwards from its equilibrium position: Resolving vertically, we have; Resultant force on mass = Force exerted by spring - Weight of mass => ma = k(e + x) - mg ------- 2 From 1, we have: ma = mg + kx - mg => a = (k/m)x Since a is proportional to displacement from equilibrium position, the oscillation is simple harmonic. So, (angular velocity)2 = (k/m) => 2pi/T = (k/m)1/2 => T = 2pi (m/k)1/2 This equation shows that the time period is proportional to the square root of the mass of the attached object.
What is the angular frequency in terms of k and m when finding the oscillation of a spring-mass system?
The angular frequency () in a spring-mass system is calculated using the formula (k/m), where k is the spring constant and m is the mass of the object attached to the spring.
