a mass of 0.30kg is attached to a spring and is set in motion with a period of 0.24 s. what is the spring constantof the spring
1 answer:
Answer:
The spring constant of the spring is 205.42 N/m.
Explanation:
Springs have their own natural "spring constants" that define how stiff they are. The letter k is used for the spring constant, and it has the units N/m.
k = -F/x
The period of a spring-mass system is proportional to the square root of the mass and inversely proportional to the square root of the spring constant.
Given:
mass of object in SHM = m = 0.30 kg
Time period of the spring mass system = 0.24s
Spring constant = k ?
Finding 'k' using Time period 'T':
We know that
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The gravitational force between two masses m₁ and m₂ is
where
G = 6.67408 x 10⁻¹¹ m³/(kg-s²), the gravitational constant
d = distance between the masses.
Given:
F = 1.5 x 10⁻¹⁰ N
m₁ = 0.50 kg
m₂ = 0.1 kg
Therefore
1.5 x 10⁻¹⁰ N = (6.67408 x 10⁻¹¹ m³/(kg-s²))*[(0.5*0.1)/(d m)²]
d² = [(6.67408x10⁻¹¹)*(0.5*0.1)]/1.5x10⁻¹⁰
= 0.0222
d = 0.1492 m = 149.2 mm
Answer: 149.2 mm
where
G = 6.67408 x 10⁻¹¹ m³/(kg-s²), the gravitational constant
d = distance between the masses.
Given:
F = 1.5 x 10⁻¹⁰ N
m₁ = 0.50 kg
m₂ = 0.1 kg
Therefore
1.5 x 10⁻¹⁰ N = (6.67408 x 10⁻¹¹ m³/(kg-s²))*[(0.5*0.1)/(d m)²]
d² = [(6.67408x10⁻¹¹)*(0.5*0.1)]/1.5x10⁻¹⁰
= 0.0222
d = 0.1492 m = 149.2 mm
Answer: 149.2 mm