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?
m/s2
To find the magnitude of the acceleration at the point of maximum extension of the spring, we'll need to use the formula for the acceleration of an object undergoing simple harmonic motion (SHM).
The formula for the acceleration in SHM is given by:
a = -ω^2 * x
Where:
a = acceleration
ω (omega) = angular frequency
x = displacement from equilibrium position
In this case, since the system is a spring-and-mass system, the angular frequency (ω) can be calculated using the formula:
ω = 2π / T
Where:
T = period of oscillation
Given:
T = 0.56 s (period)
First, we'll calculate the angular frequency (ω):
ω = 2π / T
= 2π / 0.56 s
≈ 11.18 rad/s
Next, we'll calculate the displacement (x) at the point of maximum extension. The amplitude (A) is given as 4.1 cm, and since the maximum displacement occurs at the end of the amplitude, x = A.
x = 4.1 cm
Finally, we can substitute the values into the formula for acceleration (a):
a = -ω^2 * x
= -(11.18 rad/s)^2 * 4.1 cm
Let's convert cm to meters by dividing by 100:
a ≈ -(11.18 rad/s)^2 * 0.041 m
Now we can calculate the magnitude of the acceleration by taking the absolute value:
|a| ≈ |-(11.18 rad/s)^2 * 0.041 m|
Evaluating this expression gives the magnitude of the acceleration at the point of maximum extension of the spring.