A spring balance has a scale that reads from 0to50 kg.The length of the scale is 20cm.A body suspended from this balance,when displaced & released, oscillates with period of 0.6 s.What is the weight of the body?
To calculate the weight of the body, we need to use the formula for the period of oscillation of a mass-spring system:
T = 2π * √(m/k)
Where:
T = period of oscillation
m = mass of the body
k = spring constant
In this case, we know the period of oscillation (T = 0.6 s) and the length of the scale (20 cm), but we don't know the spring constant or the mass of the body.
To find the spring constant:
1. Convert the length of the scale from cm to meters (20 cm = 0.2 m).
2. Use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position:
F = k * x
where F is the force, k is the spring constant, and x is the displacement.
Since the scale of the spring balance reads from 0 to 50 kg, we know that when the scale is at its maximum displacement of 0.2 m, it reads 50 kg. So the force exerted by the spring at maximum displacement is equal to the weight of a 50 kg object.
3. Convert the mass from kg to Newtons (Weight = Mass * gravitational acceleration = Mass * 9.8 m/s²).
So, we have:
F = k * x
50 kg * 9.8 m/s² = k * 0.2 m
k = (50 kg * 9.8 m/s²) / 0.2 m
Now that we have the spring constant, we can use the formula for the period of oscillation to find the mass of the body:
T = 2π * √(m/k)
0.6 s = 2π * √(m / [(50 kg * 9.8 m/s²) / 0.2 m])
Simplifying the equation, we can solve for the mass of the body (m):
m = [(0.6 s / (2π))^2] * [(50 kg * 9.8 m/s²) / 0.2 m]
By plugging in the values and calculating, we can find the weight of the body.