A body vibrate in s.h.m with a frequency of 50Hz and a mplitude of 4cm.find the period,the acceleration at the middle and at the end of the path oscillation,the velocity at the middle and at end of path of oscillation
5 cm = 0.05 meters
T = period = 1/50 = 0.02 second
x = 0.05 sin 2 pi t/T = 0.05 sin 314 t
v = 0.05(314) cos 314 t
a = -0.05 (314)^2 sin 314 t
Now you should be able to figure it out
A body vibrate in SHM with a frequency of 5ohertz and amplitude of 4cm. Find:-
1. the period
2. the acceleration at the middle and at the end of the path of oscillation
3. the velocities at the middle and at the end of path of oscillation
4. the velocity and acceleration at a distance 2cm from a centre of oscillation
That is not the Question
1.T=1/f =1/50=0.02s
2. The acceleration at the middle point is zero
ii). A(max)=Aw²
w=2πf
w²=4π²f²
A(max)=0.04×4×3.14²×50²
=3944m/s
Oh boy, time to do some vibrations and calculations! Let's get started.
To find the period, we'll use the formula T = 1/frequency. So, T = 1/50Hz = 0.02 seconds. Easy peasy!
To find the acceleration at the middle and the end of the path, we'll use the formula a = (2πf)^2 * amplitude.
For the middle of the path, the acceleration would be a = (2π * 50Hz)^2 * 0.04m = 784m/s². That's a nice burst of acceleration!
At the end of the path, the acceleration would be the same, since we're assuming it's undergoing simple harmonic motion. So it's also 784m/s².
Now, for the velocity at the middle and the end of the path. We'll use the formula v = 2πf * amplitude.
For the middle of the path, the velocity would be v = 2π * 50Hz * 0.04m = 12.56m/s. That's some speedy movement!
And at the end of the path, the velocity would be the same as well since it's just a simple harmonic motion. So it's also 12.56m/s.
Hope that brightened up your day with some wiggles and calculations!
To find the period (T) of a body vibrating in simple harmonic motion (SHM), we can use the formula:
T = 1 / f
where f is the frequency. In this case, the frequency is given as 50Hz. Substituting the value in the formula:
T = 1 / 50Hz
T = 0.02s
So, the period of the body's oscillation is 0.02 seconds.
To find the acceleration at the middle and at the end of the path of oscillation, we can use the formula for acceleration in SHM:
a = -ω²x
where a is the acceleration, ω is the angular frequency, and x is the displacement from the equilibrium position. The angular frequency (ω) can be calculated using the formula:
ω = 2πf
In this case, the frequency (f) is 50Hz, so:
ω = 2π * 50Hz
ω = 100π rad/s
Now, let's calculate the acceleration at the middle and at the end of the path of oscillation:
1. Acceleration at the middle (x = 0):
a = -ω²x
= -ω² * 0
= 0
So, the acceleration at the middle of the oscillation path is 0.
2. Acceleration at the end (x = amplitude):
a = -ω²x
= -ω² * amplitude
= - (100π rad/s)² * (4cm)
≈ - (100π)² * 4 cm/s²
Now, let's calculate the velocity at the middle and at the end of the path of oscillation.
The maximum velocity (v₁) of a body in SHM can be expressed as:
v₁ = ωA
where ω is the angular frequency and A is the amplitude of oscillation.
1. Velocity at the middle (x = 0):
v₁ = ω * 0
= 0
So, the velocity at the middle of the oscillation path is 0.
2. Velocity at the end (x = amplitude):
v₁ = ωA
= (100π rad/s) * (4cm)
≈ 400π cm/s
So, the velocity at the end of the oscillation path is approximately 400π cm/s.