An object perform S. H. M. of amplitude 5cm and time period 4s. If tim...
Answer:
Given:
Amplitude, A = 5cm
Time period, T = 4s
Initial displacement, x = 0
I) Frequency of oscillation:
Frequency of oscillation, f = 1/T
f = 1/4s
f = 0.25Hz
Therefore, the frequency of oscillation is 0.25Hz.
II) Displacement at 0.5s:
The displacement of the object at any given time t can be given by the equation:
x = A sin(2πft)
where A is the amplitude, f is the frequency, and t is the time.
Given that the object is at the center of the oscillation, IE, x = 0
Thus, the equation becomes:
0 = A sin(2πft)
sin(2πft) = 0
This means that the object is at the center of the oscillation at t = 0, T/2, 2T/2, 3T/2, etc.
At t = 0.5s, the object is at the extreme position, so
x = A sin(2πft)
x = 5cm sin(2π x 0.25Hz x 0.5s)
x = 5cm sin(π)
x = 0
Therefore, the displacement of the object at 0.5s is 0cm.
III) Maximum acceleration of the object:
The acceleration of the object at any given time t can be given by the equation:
a = -Aω^2sin(ωt)
where A is the amplitude, ω is the angular frequency, and t is the time.
Angular frequency, ω = 2πf
ω = 2π x 0.25Hz
ω = 1.57rad/s
At the extreme position, x = A, so
a = -Aω^2
a = -5cm x (1.57rad/s)^2
a = -12.4cm/s^2
The maximum acceleration of the object is 12.4cm/s^2.
IV) Velocity at a displacement of 3cm:
The velocity of the object at any given time t can be given by the equation:
v = -Aωcos(ωt)
where A is the amplitude, ω is the angular frequency, and t is the time.
Angular frequency, ω = 2πf
ω = 2π x 0.25Hz
ω = 1.57rad/s
At x = 3cm, the equation becomes:
3cm = 5cm sin(ωt)
sin(ωt) = 0.6
ωt = sin^-1(0.6)
ωt = 0.64 rad
At t = 0.41s, the object is at x = 3cm