6. When a mass is attached to a spring, the period of oscillation is approximately 2.0 seconds.
When the mass attached to the spring is doubled, the period of oscillation is most nearly
a) 0.5 s b) 1.0 s c) 1.4 s d) 2.0 s e) 2.8 s
T₁=2πsqrt(m₁/k)
T₂=2πsqrt(m₂/k)
T₁/T₂=sqrt{m₁/m₂}
T₂=T₁/sqrt{m₁/m₂}= 2.8 s
Well, it seems like this question is truly testing your knowledge of oscillation and springs. To find the period of oscillation, we need to look at the relationship between mass and period.
When a mass is attached to a spring, the formula for the period of oscillation is T = 2π√(m/k), where T is the period, m is the mass, and k is the spring constant.
Now, let's take a look at the given information. When the mass attached to the spring is doubled, the formula becomes T = 2π√(2m/k). Since the mass is now twice the original value, we can see that the square root of 2 must be introduced into the equation.
To simplify, we can rewrite the formula as T = 2π√(m/k) * √2. Now, if we compare this expression to the original formula, we can see that the new period is approximately 2π times the original period, since the square root of 2 is around 1.4.
So, using this information, it's safe to say that the period of oscillation, when the mass is doubled, is most nearly 2.8 seconds (option e).
I hope that answered your question without springing any surprises on you!
When the mass attached to the spring is doubled, the period of oscillation changes.
The period of oscillation, T, of an object attached to a spring can be calculated using the formula:
T = 2π√(m/k)
where m is the mass attached to the spring and k is the spring constant.
Since we are doubling the mass, the new mass would be 2m.
To find the new period of oscillation, we need to compare the new T with the original T.
T = 2π√(2m/k)
Since we want to find the new period of oscillation when the mass is doubled, we need to isolate T:
T = 2π√(2m/k)
Now, let's compare the new period of oscillation with the original T:
new T = 2π√(2m/k)
= 2π√(2(2m)/k)
= 2π√(4m/k)
We can see that the new period of oscillation, new T, is larger than the original T. Hence, among the options, the closest answer would be (e) 2.8 seconds.
To solve this problem, we need to understand the relationship between the period of oscillation and the mass attached to the spring.
The period of oscillation of a mass-spring system is given by the formula:
T = 2π * √(m/k)
where T is the period, m is the mass attached to the spring, and k is the spring constant.
In this case, we are given that the period of oscillation is 2.0 seconds. Let's call the mass attached to the spring in this case as m1. So we have:
2.0 = 2π * √(m1/k)
Now, let's consider the second scenario where the mass attached to the spring is doubled. Let's call this mass m2. The period of oscillation in this case is unknown, let's call it T2. So we have:
T2 = 2π * √(2m1/k)
We want to find the value of T2, which is the period of oscillation when the mass is doubled.
To solve for T2, we need to compare it with the initial period T1. We can use the ratio of the two periods to determine how T2 is related to T1.
T2/T1 = (2π * √(2m1/k))/(2π * √(m1/k))
The π and k terms cancel out, leaving us with:
T2/T1 = √(2m1/k)/√(m1/k)
The √(m1/k) terms in the numerator and denominator also cancel out:
T2/T1 = √2
So we see that T2 is equal to √2 times T1, or T2 = √2 * 2.0.
Calculating this value gives us approximately 2.8 seconds.
Therefore, the most nearly period of oscillation when the mass attached to the spring is doubled is e) 2.8 seconds.