Physics

Spring Mass Systems

172 Questions

A spring mass system is a key physics concept used to study oscillations and simple harmonic motion. It involves understanding spring constants, damping, and series or parallel combinations. These principles are frequently tested in engineering entrance examinations.

Series and parallel springsSpring constant calculationsDamped oscillationsSpring compression energyElevator systems

Spring Mass Systems Questions

Multiple choice
  1. Boyle's law

  2. Hooke's law

  3. Bernouli's theorem

  4. Pascal's law

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Hooke's law states that the extension of a spring is directly proportional to the applied force, which is the principle behind spring balances.

Multiple choice
  1. Mass is measured with a spring balance and weight is measured with a pan balance.

  2. Mass is measured with a pan balance and weight is measured with a spring balance.

  3. Both the mass and weight are measured with a spring balance.

  4. Both the mass and weight are measured with a pan balance.

  5. Both 1 and 2.

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Mass is the quantity of matter in a body regardless of its volume or of any forces acting on it. Mass is measured using a pan balance, a triple-beam balance, lever balance or electronic balance. While weight is a measurement of the gravitational force acting on an object. It is measured using a spring balance. So, it is incorrect

Multiple choice
  1. 1275 m/s; 60 Hz

  2. 1875 Hz; 44 m/s

  3. 1875 m/s; 44 Hz

  4. 4113 m/s; 60 Hz

Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Seismic mass = (M) 0.06 kg Spring constant (K) = 4500 N/m (Natural angular velocity)2 = K/M = 4500/0.06 = 273.86 rad /sec Maximum mass displacement (A) = 0.025 m Maximum acceleration = (Natural angular velocity)2 x maximum mass displacement = (273.86)2 x 0.025 = 1875 m/s Natural Frequency (f0) = Natural angular velocity/2 pi = 44 Hz

Multiple choice
  1. R1 = 20 kN and R2 = 40 kN

  2. R1 = 50 kN and R2 = 50 kN

  3. R1 = 30 kN and R2 = 60 kN

  4. R1 = 40 kN and R2 = 8O kN

Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

 $\text{If} P = 100KN \\ \text{Spring force at point (1)} k \delta_1 = R_1 = \frac{2P}{5} = 40 KN\\ \text{Spring force at point (2)} k \delta_2 = R_2 = \frac{4P}{5} = 80 KN$