Question 18:
Let the spring constant is k
when an additional 40 g mass is attached to the spring.
The change in reading = 0.599 - 0.55 = 0.049 m
Now, k * x = m * g
k * 0.049 = 0.040 * 9.8
k = 8 Nm^-1
Qusetion 19:
If 34 g is added.
extra extension in spring = m * g/k
extra extension in spring = 0.034*9.8 / 8
extra extension in spring =0.042 m = 4.2 cm
Question 20:
As the motion will be simple harmonic
frequency = 1/(2pi) * sqrt(k/total mass)
frequency = 1/(2pi) * sqrt(8/(0.100 + 0.040 + 0.034)) = 1/2pi*6.780 = 0.023 Hz
frequency = 0.023 Hz
Question 18 Not compiete Marked out of 100 stripy rectangle in the second part of the...
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