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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:
#SPJ4
Answer:
b) k Δx - W cos θ - μ mg cos θ = m a , c) θ = 86.6º, d) Δx = 1.18 m
Explanation:
a) In the attachment we can see a diagram of the forces in this problem and the coordinate axes for its decomposition.
F is the force applied by the spring, while it is compressed, this force disappears when the block leaves the spring
b) Let's apply Newton's second law for when the spring is compressed
let's use trigonometry to break down the weight
sin θ = Wₓ / W
cos θ = W_y / W
Wₓ = W sin θ
W_y = W cos θ
Y axis
N - W_y = 0
N = W_y
N = W cos θ
X axis
F -Wₓ -fr = ma
the force applied by the spring is given by hooke's law
F = k Δx
friction force has the expression
fr = μ N
fr = μ W cos θ
we substitute
k Δx - W cos θ - μ mg cos θ = m a ( 1)
c) If the plane has no friction, what is the angle so that Δx = 0.1m
We write the equation 1, with fr = 0 and since the system is still a = 0
k Δx - W cos θ -0 = 0
cos θ =
cos θ =
cos θ = 0.0598
θ = cos⁻¹ 0.0598
θ = 86.6º
d) In this part they give the angle θ = 45º and there is no friction, they ask the compression
the acceleration is zero, we substitute in 1
k Δx - W cos θ - 0 = 0
Δx =
Δx =
Δx = 1.18 m
