A body vibrates in SHM (simple harmonic motion) with the frequency of 50HZ and an amplitude of 4cm, find the (a) the period (b) the acceleration at the middle and the end of the path of oscillation.?
the period is 1/frequency, right?
The period of sin(kx) is 2pi/k
now take derivatives for acceleration
Solve it now
To find the period of the simple harmonic motion (SHM), you can use the formula:
Period (T) = 1 / Frequency (f)
Given that the frequency is 50 Hz, we can find the period as follows:
T = 1 / 50 Hz
T = 0.02 seconds
Therefore, the period of the motion is 0.02 seconds.
Now, let's move on to finding the acceleration at the middle and the end of the path of oscillation.
In SHM, acceleration can be expressed as:
Acceleration (a) = -ω^2 * x
Where:
- ω is the angular frequency, given by ω = 2πf
- x is the displacement from the equilibrium position
Given that the frequency is 50 Hz, we can calculate the angular frequency as follows:
ω = 2π * 50 Hz
ω = 100π rad/s
At the middle of the path of oscillation, the displacement (x) is zero. Therefore, the acceleration at the middle can be calculated as:
a = -ω^2 * x
a = - (100π rad/s)^2 * 0
a = 0
Since the displacement is zero, the acceleration at the middle is zero.
At the end of the path of oscillation (amplitude), the displacement (x) is equal to the amplitude, which is 4 cm (or 0.04 m). Therefore, the acceleration at the end can be calculated as:
a = -ω^2 * x
a = - (100π rad/s)^2 * 0.04 m
Now, you can calculate the acceleration at the end of the path of oscillation using the above formula and evaluate the numerical value.