Answer:
Part A)
A = 10 cm
Part b)
x = 10 cm
Explanation:
As we know that time period of oscillations for spring block system is given by the equation
[tex]T = 2\pi \sqrt{\frac{m}{k}}[/tex]
here we know that
[tex]T = 1.5 s[/tex]
Part A)
we know that the maximum speed of the block at mean position is given as
[tex]v_{max} = A\omega[/tex]
[tex]v_{max} = 42 cm/s[/tex]
[tex]v_{max} = A\frac{2\pi}{T}[/tex]
[tex]0.42 = A\frac{2\pi}{1.5}[/tex]
[tex]A = 0.10 m = 10 cm[/tex]
Part B)
As we know that glider starts from its maximum position
so here we have
[tex]x = A cos\omega t[/tex]
[tex]x = (10 cm)cos(\frac{2\pi}{T} t)[/tex]
[tex]x = (10 cm) cos(\frac{2\pi}{1.5}\times 30)[/tex]
[tex]x = 10 cm[/tex]
The amplitude is 10cm and the position of the glider is 10cm.
The period of oscillations for spring block system is given below.
T=2π√m/k
T=1.5x.
T is time.
m is mass
k is constant.
Therefore,the maximum speed of the block at mean position is given below.
Vmax=Aw
where
V max is maximum velocity
A is amplitude
Vmax=0.42
Vmax=A2π/T
0.422=A2π/1.5
A=0.1m
The amplitude is 10cm.
The glider's position at time 30s.
As we know that glider starts from its maximum position.
x=Acoswt
where A is amplitude
x is distance
x=10×cos(2πt/T
x=10×cos2π/1.5×30
x=10cm.
Therefore, The amplitude is 10cm and the distance is 10cm.
Oscillatory motion is refers to the movement of body to and fro motion from a mean position. The ideal condition is that the object can be in oscillatory motion .
Therefore, The amplitude is 10cm and the distance is 10cm..
Learn more on Oscillatory motion from here.
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