A body is supported by a spiral spring and causes a stretch of 1.5cm in the spring. If the mass is now set in vertical oscillation of small amplitude, what is the periodic time of oscillation
Doesn't angular frequency= sqrt (k/m) ?
where k= mg/.0015
or 2PI/T= sqrt (g/.0015)
solve for period T.
http://hyperphysics.phy-astr.gsu.edu/hbase/shm.html
Yes, you are correct. The angular frequency (ω) of an oscillating system is given by ω = √(k/m), where k is the spring constant and m is the mass of the body.
In this case, you have mentioned that the stretch of the spring is 1.5 cm, which we can convert to meters by dividing by 100. So, the stretch of the spring in meters is 0.015 m.
To find the spring constant (k), we can use the equation k = mg/Δx, where m is the mass of the body, g is the acceleration due to gravity, and Δx is the stretch of the spring.
Using the given values, k = mg/Δx = m * 9.8 / 0.015.
Now, we can substitute the calculated value of k into the equation for angular frequency: ω = √(k/m) = √((m * 9.8 / 0.015) / m) = √(9.8 / 0.015) = √(653.33).
Finally, to find the period of oscillation (T), we can use the equation 2π/T = ω. Rearranging the equation, we have T = 2π/ω.
Substituting the calculated value of ω, we get T = 2π/√(653.33), which simplifies to T ≈ 0.0354 seconds (s).
Therefore, the periodic time of oscillation for the given system is approximately 0.0354 seconds.