a spring stretches by 3.90 cm when a 10gm mass is hung from it. if a 25 gm mass attached to this spring oscillates in simple harmonic motion, calculate the period of the oscillation
To find the period of the oscillation, we can use Hooke's Law and the formula for the period of a mass-spring system. Here's how to do it:
1. Start by calculating the spring constant, denoted by 'k'. According to Hooke's Law, the force exerted by a spring is proportional to its displacement. In this case, with a 10g mass stretching the spring by 3.90 cm (which is 0.039 m), we can write the equation as follows:
F = -k * x
Where F is the force applied by the mass, k is the spring constant, and x is the displacement of the spring.
Since the force applied is the weight of the mass (10g), we can convert it to Newtons using the equation:
Force = mass * acceleration due to gravity
Mass = 10g = 0.01 kg (given that 1 kg = 1000 g)
Acceleration due to gravity (g) = 9.8 m/s^2
Plugging these values into the equation:
Force = 0.01 kg * 9.8 m/s^2 = 0.098 N
Therefore:
0.098 N = -k * 0.039 m
2. Rearrange the equation to solve for 'k':
k = -0.098 N / 0.039 m ≈ -2.513 N/m
The negative sign indicates that the force exerted by the spring is in the opposite direction of the displacement.
3. Now we can use the formula for the period of a mass-spring system:
T = 2π * sqrt(m / k)
Where T is the period of the oscillation, m is the mass attached to the spring, and k is the spring constant.
Plugging in the values:
m = 25g = 0.025 kg
T = 2π * sqrt(0.025 kg / -2.513 N/m)
Performing the calculations:
T ≈ 2 * 3.14159 * sqrt(0.025 kg / -2.513 N/m)
T ≈ 2 * 3.14159 * sqrt(0.009952 kg/m)
T ≈ 2 * 3.14159 * 0.09976 s ≈ 0.626 s
Therefore, the period of the oscillation is approximately 0.626 seconds.