Question

1. The potential energy of an object attached to a spring is 2.80 J at a location where the kinetic energy is 1.50 J.1.50 J. If the amplitude ?A of the simple harmonic motion is 20.0 cm, calculate the spring constant k and the magnitude of the largest force Fspring, max that the object experiences.

k = ??? N/m

F spring, max = ??? N

2. The spaceship Intergalactica lands on the surface of the uninhabited Pink Planet, which orbits a rather average star in the distant Garbanzo Galaxy. A scouting party sets out to explore. The party's leader–a physicist, naturally–immediately makes a determination of the acceleration due to gravity on the Pink Planet's surface by means of a simple pendulum of length 1.37 m. She sets the pendulum swinging, and her collaborators carefully count 104 complete cycles of oscillation during 208 s. What is the result?

acceleration due to gravity = ??

Answer 1

A)     K.E+ P.E = total energy

1.5+2.8= 1/2kA^2

4.2×2/A^2=k

k= 210N/m

F= - kx

For max force F= - kA

F= - 210× 0.20= 42N

2)

The expression for the time period of the pendulum is.

T = 2\pi \sqrt {\frac{L}{g}}T=2πgL​​

Squaring above equation on both sides.

{T^2} = 4{\pi ^2}\frac{L}{g}T2=4π2gL​

Rewrite the above expression for g as follows:

The acceleration due to gravity is directly proportional to the length of the pendulum and inversely proportional to the square of the time period.

Substitute 2.07 s for T and 1.10 m for L in the equation g = \frac{{4{\pi ^2}L}}{{{T^2}}}g=T24π2L​.

\begin{array}{c}\\g = \frac{{4{\pi ^2}\left( {1.10\;{\rm{m}}} \right)}}{{{{\left( {2.07\;{\rm{s}}} \right)}^2}}}\\\\ = 22.29\;{\rm{m/}}{{\rm{s}}^2}\\\end{array}g=(2.07s)24π2(1.17m)​=22.29m/s2​

Therefore, the acceleration due to gravity on the pink planet’s surface is 22.29\;{\rm{m/}}{{\rm{s}}^2}10.13m/s2.