Answer:
A=0.199
Explanation:
We are given that
Mass of spring=m=450 g=[tex]=\frac{450}{1000}=0.45 kg[/tex]
Where 1 kg=1000 g
Frequency of oscillation=[tex]\nu=1.2Hz[/tex]
Total energy of the oscillation=0.51 J
We have to find the amplitude of oscillations.
Energy of oscillator=[tex]E=\frac{1}{2}m\omega^2A^2[/tex]
Where [tex]\omega=2\pi\nu[/tex]=Angular frequency
A=Amplitude
[tex]\pi=\frac{22}{7}[/tex]
Using the formula
[tex]0.51=\frac{1}{2}\times 0.45(2\times \frac{22}{7}\times 1.2)^2A^2[/tex]
[tex]A^2=\frac{2\times 0.51}{0.45\times (2\times \frac{22}{7}\times 1.2)^2}=0.0398[/tex]
[tex]A=\sqrt{0.0398}=0.199[/tex]
Hence, the amplitude of oscillation=A=0.199
The amplitude of oscillation will be "0.199".
Given:
Mass of spring,
or,
= 0.45 kg
Frequency,
Total energy,
As we know the formula,
→ [tex]E = \frac{1}{2} m \omega^2 A^2[/tex]
By putting the values, we get
→ [tex]0.51= \frac{1}{2}\times 045 (2\times \frac{22}{7}\times 1.2 )^2 A^2[/tex]
→ [tex]A^2 = \frac{2\times 0.51}{0.45(2\times \frac{22}{7}\times 1.2 )^2}[/tex]
[tex]= 0.0398[/tex]
→ [tex]A = \sqrt{0.0398}[/tex]
[tex]= 0.199[/tex]
Thus the above response is right.
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