A maass of 0.5kg is attached to one end of the helical spring and produces an extension 2.5cm the mass now set into vertical oscillation of amplitude 10mm the period of oscillation is??....g=10ms2
To find the period of oscillation, we can use Hooke's Law and the formula for the period of a mass-spring system.
First, let's determine the spring constant of the helical spring using Hooke's Law. Hooke's Law states that the force exerted by a spring is directly proportional to the extension or compression of the spring.
F = k * x
Where:
F = Force exerted by the spring (in Newtons)
k = Spring constant (in Newtons per meter)
x = Extension or compression of the spring (in meters)
In this case, the mass attached to the spring produces an extension of 2.5 cm (or 0.025 m). The mass is given as 0.5 kg.
We can use Newton's second law to find the force exerted by the mass:
F = m * g
Where:
m = Mass (in kg)
g = Acceleration due to gravity (in m/s^2)
Plugging in the values, we have:
F = 0.5 kg * 10 m/s^2
F = 5 N
Now, equating the force from Hooke's Law and the force from the mass, we have:
k * x = m * g
Plugging in the values:
k * 0.025 m = 0.5 kg * 10 m/s^2
Simplifying the equation, we find:
k = (0.5 kg * 10 m/s^2) / 0.025 m
k = 20 N/m
Now, we can use the formula for the period of a mass-spring system:
T = 2π * √(m / k)
Plugging in the values:
T = 2π * √(0.5 kg / 20 N/m)
Calculating the period:
T ≈ 2π * √(0.025 kg / N)
T ≈ 2π * √0.00125 kg/N
T ≈ 2π * 0.035
Finally, we get:
T ≈ 0.22 seconds
Therefore, the period of oscillation is approximately 0.22 seconds.