Classical Mechanics

Simple Harmonic Motion

Solutions to Simple Harmonic Motion

         

A 40 g40\text{ g} cube of edge length l=3 cml=3\text{ cm} floats on water, oscillating up and down. Initially at t=0,t=0, the position of the body was x(t=0)=5 cm  x(t=0) = 5 \text{ cm } and the velocity was v(t=0)=0 m/s , v(t=0) = 0 \text{ m/s }, where xx denotes the vertical difference in the position of the cube from the equilibrium point x0.x_0. What is the velocity v(t)v(t) of the body at t=2 s ? t = 2 \text{ s } ?

Assumptions and Details

  • Positive sign of x(t)x(t) means downward direction.
  • Ignore any friction.
  • The density of water is ρ=1 g/cm3. \rho = 1 \text{ g/cm}^3.
  • The gravitational acceleration is g=10 m/s2.g=10\text{ m/s}^2.
Show explanation
View wiki
by Brilliant Staff

As illustrated in the above diagram, a body of mass m=1 kgm = 1 \text{ kg} attached to a spring with spring constant k=9 N/m k = 9 \text{ N/m} is oscillating on a frictionless floor. Initially at t=0,t=0, the position of the body was x(t=0)=1 m  x(t=0) = 1 \text{ m } and the velocity was v(t=0)=0 m/s , v(t=0) = 0 \text{ m/s }, where xx denotes the difference in the length of the spring from its original length.

What is the position of the body at t=3 s ? t = 3 \text{ s } ?

Show explanation
View wiki
by Brilliant Staff

A 40 g40\text{ g} cube of edge length l=5 cml=5\text{ cm} floats on water, oscillating up and down. Initially at t=0,t=0, the position of the body was x(t=0)=4 cm  x(t=0) = 4 \text{ cm } and the velocity was v(t=0)=0 m/s , v(t=0) = 0 \text{ m/s }, where xx denotes the vertical difference in the position of the cube from the equilibrium point x0.x_0. What is x(t)x(t) at t=2 s ? t = 2 \text{ s } ?

Positive sign of x(t)x(t) means downward direction.
Ignore any friction.
The density of water is ρ=1 g/cm3. \rho = 1 \text{ g/cm}^3.
The gravitational acceleration is g=10 m/s2.g=10\text{ m/s}^2.

Show explanation
View wiki
by Brilliant Staff

As illustrated in the above diagram, a body of mass m=3 kgm = 3 \text{ kg} attached to a spring with spring constant k=75 N/m k = 75 \text{ N/m} is oscillating on a frictionless floor. Initially at t=0,t=0, the position of the body was x(t=0)=1 m  x(t=0) = 1 \text{ m } and the velocity was v(t=0)=0 m/s , v(t=0) = 0 \text{ m/s }, where xx denotes the difference in the length of the spring from its original length. What is the velocity of the body at t=5 s ? t = 5 \text{ s } ?

Show explanation
View wiki
by Brilliant Staff

A body of mass m=6 kgm = 6 \text{ kg} is oscillating on two identical springs with spring constant k=48 N/m, k = 48 \text{ N/m}, as shown in the above diagram, where xx denotes the downward difference from the equilibrium point x0.x_0. Initially, the difference of the body was x(t=0)=4 m  x(t=0) = 4 \text{ m } and the velocity was v(t=0)=0 m/s . v(t=0) = 0 \text{ m/s }. What is the position of the body at t=4 s ? t = 4 \text{ s } ?

Assumptions and Details

  • The gravitational acceleration is g=10m/s2. g = 10 \text{m/s}^2.
Show explanation
View wiki
by Brilliant Staff
×

Problem Loading...

Note Loading...

Set Loading...