A mass m is attached to an ideal massless spring with spring constant k. In experiment 1 the mass oscillates with amplitude a, a
1 answer:
Time period remains the same in both the experiment as change in amplitude does not affect time period.
What are the factors on which time period depends in SHM?
Time period is given by:
where,
T = time period
m = mass
k = spring constant
In a straightforward harmonic motion, we see from the preceding formula that the time period depends only on the object's mass and spring constant (SHM). The time period will adjust to any variations in the object's mass or the spring constant.
What is Spring Constant?
A spring's "spring constant" is a property that quantifies the relationship between the force acting on the spring and the displacement it produces. In other words, it characterises a spring's stiffness and the extent of its range of motion.
Learn more about SHM here:
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Answer:
The minimum average speed the opponent must move so that he is in position to hit the ball is approximately 5.79 m/s
Explanation:
The given parameters of the ball are;
The initial speed of the ball = 15 m/s
The direction in which the ball is launched = 50° above the horizontal
The location of the other tennis player when the ball is launched = 10 m from the ball
The time at which the other tennis player begins to run = 0.3 seconds after the ball is launched
The height at which the ball is hit back = 2.1 m above the height from which the ball is launched
The vertical position, 'y', at time, 't', of a projectile motion is given as follows;
y = (u·sinθ)·t - 1/2·g·t²
When y = 2.1 m, we have;
2.1 = (15·sin(50°))·t - 1/2·9.8·t²
∴ 4.9·t² - (15·sin(50°))·t + 2.1 = 0
Solving with the aid of a graphing calculator function, we get;
t = 0.199776187257 s or t = 2.14525782198 s
Therefore, the ball is at 2.1 m above the start point on the other side of the court at t ≈ 2.145 seconds
The horizontal distance, 'x', the ball travels at t ≈ 2.145 seconds is given as follows;
x = u × cos(50°) × t = 15 × cos(50°) × 2.145 ≈ 20.682 m
The horizontal distance the ball travels at t ≈ 2.145 seconds, x ≈ 20.682 m
Therefore, we have;
The time the other player has to reach the ball, t₂ =2.145 s - 0.3 s ≈ 1.845 s
The distance the other player has to run, d = 20.682 m - 10 m = 10.682 m
The minimum average speed the other player has to move with, = d/t₂
∴ = 10.682 m/(1.845 s) ≈ 5.78970189702 m/s ≈ 5.79 m/s
The minimum average speed the opponent must move so that he is in position to hit the ball, ≈ 5.79 m/s.