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2 answers:
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When the mass of the spring changed from 0.2kg to 0.1kg, the time period changed from 1 sec to 0.5 seconds
<u>Explanation:</u>
Given-
Mass, m1 = 0.2kg
Time period, T1 = 1s
m2 = 0.1 kg
T2 = ?
We know,
where,
T = Time period
m = mass
k = spring constant
From the equation, we can see that T is directly proportion to the square root of mass, m
T ∝ √m
So,
If m1 = 0.2kg , T1 = 1s and m2 = 0.1kg
The T2 would be:
Therefore, when the mass of the spring changed from 0.2kg to 0.1kg, the time period changed from 1 sec to 0.5 seconds
Answer:
A). 2.3 s
Explanation:
Given that,
Length of the string = 130cm or 1.3 m
g near Earth's surface = 9.8
To determine the period of a pendulum near Earth’s surface;
T = 2 π
T = 2π
T = 2.3 sec
Therefore, the period of the pendulum near the earth's surface is 2.3 sec.