A block of mass 200 grams is connected to a light, horizontal spring with a spring constant k= 5 N/m. It is free to oscillate on a frictionless surface, and we’ll also neglect air resistance so you don’t have to worry about decaying. If the block is displaced 5 cm away from equilibrium and released from rest, compute the period Tof its motion.
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Ответ:
T = 1.26s
Explanation: From the question,
Mass of loaded spring = 200g = 0.2
Spring constant of spring = 5 N/m
Amplitude = 5cm = 0.05m
The angular frequency (ω) of the harmonic motion of a loaded spring is related to spring constant (k) and mass of loaded spring (m), this is given below as
ω = √k/m
ω = √5/0.2
ω = √25
ω = 5 rad/s
Recall that for a simple pendulum, ω = 2πf where f = frequency of oscillations.
Hence f = ω/2π.
But T = 1/f where T = period
Hence T = 2π/ω
T = 2 × 3.142/ 5
T = 1.26s
Ответ:
T = 1.26 seconds
The period T of its motion is 1.26 seconds
Explanation:
Given:
Mass of block m = 200grams = 0.2kg
Spring constant k = 5N/m
The period P of motion of the spring with mass, can be written as;
T = 2π√(m/k) 1
Where the values of m and k are stated above.
Substituting the values of m and k, we have;
T = 2π√(0.2/5)
T = 2π(0.2)
T = 1.26 seconds
The period T of its motion is 1.26 seconds
Ответ:
Density and boiling point
Explanation:
There are properties of matter that do not change regardless of how much of the substance you have. The boiling point of a cup of water, for example is te same as the full kettle of water. These are called intensive properties. Examples include, but are not limited to, density , specific heat capacity , hardness, boiling point , color etc