The period of oscillation of a spring-and-mass system is 0.56 s and the amplitude is 4.1 cm. What is the magnitude of the acceleration at the point of maximum extension of the spring?
To find the magnitude of the acceleration at the point of maximum extension of the spring, we need to know the angular frequency, which is the rate at which the system oscillates.
The angular frequency (ω) can be calculated using the formula: ω = 2π / T, where T is the period of oscillation.
In this case, the period is given as 0.56 s, so we can substitute the value into the formula to find the angular frequency: ω = 2π / 0.56 ≈ 11.18 rad/s.
The maximum extension of the spring occurs when the displacement is equal to the amplitude, which is given as 4.1 cm.
The acceleration (a) at any given point in a simple harmonic motion can be calculated using the formula: a = -ω²x, where x is the displacement from the equilibrium position.
For the maximum extension of the spring, x = 4.1 cm = 0.041 m, and we already found ω to be 11.18 rad/s.
Plugging these values into the formula, we get: a = -(11.18)^2 * 0.041 ≈ -19.64 m/s².
The magnitude of acceleration is the absolute value of the calculated value, so the magnitude of the acceleration at the point of maximum extension of the spring is approximately 19.64 m/s².