A toy is undergoing SHM on the end of a horizontal spring with force constant 310 N/m . When the toy is 0.120 m from its equilibrium position, it is observed to have a speed of 3 m/s and a total energy of 3.6 J . 1.) Find the mass of the toy.
2.) Find the amplitude of the motion.
3.) Find the maximum speed obtained by the object during its motion.

1) the mass of the toy is 0.304 kg

2) the amplitude of the motion is 0.1524 m

3) the maximum speed obtained by the object during its motion is 4.87 m/s

Explanation:

Given the data in the question;

force constant of spring k = 310 N/m

position of toy x = 0.120 m

speed v = 3 m/s

Total energy E = 3.6 J

let m represent mass of toy.

1.) Find the mass of the toy.

we know that; The total energy of the system equals the sum of kinetic energy of the toy and the potential energy of the spring;

E = kx² + mv²

we substitute

3.6 = ( × 310 × (0.120)² ) + ( × m × (3)² )

3.6 = 2.232 + 4.5m

3.6 - 2.232 = 4.5m

1.368 = 4.5m

m = 1.368 / 4.5

m = 0.304 kg

Therefore, the mass of the toy is 0.304 kg

2) Find the amplitude of the motion.

we know that;

E = kA²

where A is the amplitude of the motion,

we substitute

3.6 = × 310 × A²

3.6 = 155 × A²

A² = 3.6 / 155

A² = 0.0232258

A = √0.0232258

A = 0.1524 m

Therefore, the amplitude of the motion is 0.1524 m

3) Find the maximum speed obtained by the object during its motion;

we know that;

E = m²

where is the maximum speed

so we substitute

3.6 = × 0.304 × ²

3.6 = 0.152 × ²

² = 3.6 / 0.152

² = 23.6842

² = √23.6842

² = 4.8664 ≈ 4.87 m/s

Therefore, the maximum speed obtained by the object during its motion is 4.87 m/s

where are the vectors ?

measuring things 1 litre of water