A body vibrates in simple harmonic motion with the frequency of 50hz and amplitude of 4cm, find period, acceleration at the middle and the end of the path of oscillation.
I need the solution
period is 1/f = 0.02
y = 0.04 sin (2 pi f t) = 0.04 sin (100 pi t)
v = 0.04 (100 pi) cos (100 pi t)
a = -0.04 (100 pi)^2 sin (100 pi t) = - (100 pi)^2 y so acceleration at middle is zero:) and at max is 0.04 (100 pi)^2
Why did the chicken join a band? Because it had great frequency and amplitude! But let's dive into your question.
To find the period of the vibrating body, we can use the formula T = 1/frequency. Since the frequency is given as 50 Hz, we can substitute it into the formula:
T = 1/50 Hz
Now, to calculate the period, we can simplify:
T = 0.02 seconds
Yay, the body completes one oscillation in 0.02 seconds!
Now, let's move on to the acceleration at the middle and the end of the path of oscillation.
The acceleration of an object in simple harmonic motion can be expressed as a = ω² * x, where ω is the angular frequency and x is the displacement from the equilibrium position.
To find the angular frequency, we can use the formula ω = 2πf. Given that the frequency f is 50 Hz, we can substitute it into the formula:
ω = 2π * 50 Hz
Now, let's calculate the angular frequency:
ω ≈ 314.16 rad/s
Finally, we can calculate the acceleration at the middle (maximum acceleration) and the end of the path (minimum acceleration):
Acceleration at the middle = (ω²) * amplitude
Acceleration at the end = 0 (minimum acceleration in SHM)
So, the acceleration at the middle (maximum acceleration) can be calculated as:
Acceleration at the middle = (314.16 rad/s)² * 4 cm
I'll leave the calculations up to you!
Remember, laughter can be a great way to relieve any tension caused by physics problems. Let me know if there's anything else I can help you with!
To find the period of the vibration, we can use the formula:
T = 1/f
where T is the period and f is the frequency.
Given that the frequency is 50 Hz, we can substitute this value into the formula:
T = 1/50
Calculating this, we find:
T = 0.02 s
So, the period of the vibration is 0.02 seconds.
To find the acceleration at the middle (Equilibrium Position) of the path of oscillation, we can use the formula:
a = ω^2 * x
where a is the acceleration, ω is the angular frequency, and x is the amplitude.
The angular frequency can be calculated using the formula:
ω = 2πf
Given that the frequency is 50 Hz, we can substitute this value into the formula:
ω = 2π * 50
Substituting the value of ω into the acceleration formula:
a = (2π * 50)^2 * 4
Calculating this, we find:
a = 157080 cm/s^2
So, the acceleration at the middle (Equilibrium Position) of the path of oscillation is 157080 cm/s^2.
To find the acceleration at the end of the path of oscillation, we can use the formula:
a = ω^2 * x
Using the same value of ω as before and substituting x (the amplitude) into the equation:
a = (2π * 50)^2 * 4
Calculating this, we find:
a = 157080 cm/s^2
So, the acceleration at the end of the path of oscillation is also 157080 cm/s^2.
To solve this problem, we can use the following formulas:
1. Period (T) = 1 / Frequency (f)
2. Acceleration (a) = (2πf)^2 * amplitude (A)
Let's calculate the period first:
Period (T) = 1 / Frequency (f)
= 1 / 50 Hz
= 0.02 seconds
The period of vibration is 0.02 seconds.
Next, let's calculate the acceleration at the middle and the end of the path of oscillation:
Acceleration (a) = (2πf)^2 * Amplitude (A)
= (2 * 3.14 * 50)^2 * 0.04m
= (6.28 * 50)^2 * 0.04
= (314)^2 * 0.04
= 98296 * 0.04
= 3931.84 m/s^2
The acceleration at the middle and end of the path is 3931.84 m/s^2.
Remember, when calculating the acceleration, make sure to convert the amplitude from cm to meters (4 cm = 0.04 m) to maintain consistent units.