A 450g mass on a spring is oscillating at 1.2Hz. The totalenergy of the oscillation is 0.51J. What is the amplitude.
1 answer:
Answer:
A=0.199
Explanation:
We are given that
Mass of spring=m=450 g=
Where 1 kg=1000 g
Frequency of oscillation=
Total energy of the oscillation=0.51 J
We have to find the amplitude of oscillations.
Energy of oscillator=
Where =Angular frequency
A=Amplitude
Using the formula
Hence, the amplitude of oscillation=A=0.199
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Answer:
Explanation:
Given that:
For a first order instrument with a sensitivity of .4 mV/K
constant c = 25 ms = 25 × 10⁻³ s
The initial temperature = 273 K
The final temperature = 473 K
The initial volume = 0.4 mV/K × 273 K = 109.2 V
The final volume = 0.4 mV/K × 473 K = 189.2 V
the instrument’s response as a function of time for a sudden temperature increase can be computed as follows:
Let consider y to be the function of time i.e y(t).
So;
y(t) = 109.2 + (189.2 - 109.2)( 1 - )mV
y(t) = (109.2 + 80 ( 1 - )) mV
Plot the response y(t) as a function of time.
The plot of y(t) as a function of time can be seen in the diagram attached below.
What are the units for y(t)?
The unit for y(t) is mV.
Find the 90% rise time for y(t90) and the error fraction,
The 90% rise time for y(t90) is as follows:
Initially 90% of 189.2 mV = 0.9 × 189.2 mV
= 170.28 mV
170.28 mV = (109.2 + 80 ( 1 - )) mV
170.28 mV - 109.2 mV = 80 ( 1 - )) mV
61.08 mV = 80 ( 1 - )) mV
0.7635 mV = ( 1 - )) mV
t = 1.44 × 25 × 10⁻³ s
t = 0.036 s
t = 36 ms
The error fraction =
The error fraction = 0.1
The error fraction = 10%
Answer:d
Explanation:
Spring is compressed to a distance of x from its equilibrium position
Work done by block on the spring is equal to change in elastic potential energy
i.e. Work done by block
therefore spring will also done an equal opposite amount of work on the block in the absence of external force
Thus work done by spring on the block
Thus option d is correct