A steel ball is hung from a vertical ideal spring where it oscillates in simple harmonic motion with an amplitude of 0.157 m and an angular frequency of π rad/s.  Also, x(0) = 157m. Write the acceleration, in m/s2, of the ball as a function of time.

Q: There is a mass-spring that oscillates with simple harmonic motion. It has a mass of .5 kg, a period…

A: Solution:-  Time period oscillation= 2π (M/K)0.5 Where, M = mass of block.  K = spring constant.…

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Q: A mass m = 2.55 kg is at the end of a horizontal spring on a frictionless horizontal surface. The…

A: Mass of object (m) = 2.55 kg Amplitude of motion (A) = 2.5 cm = 0.025 m Frequency (f) = 1.05 Hz

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A: we are authorized to answer three sub parts at a time only.Since you have not written which part i…

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A: Given y = 1.65 sin (0.470πt) We have to calculate (a) Find the amplitude.(b) Find the period.(c)…

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A: Given : Mass attached, m = 2.75 kg Amplitude of oscillation, A = 4.5 cm = 0.045 m Frequency of…

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A: Write the given values with suitable variables. m=5 kθ=π6ω=9 rad/s Here, m denotes the mass, θ…

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Q: We have a simple pendulum, composed of a mass of 5.00kg hanging from a thread of negligible mass.…

A: Solution:   Given Values,   Mass(m)=5 Kg Angular displacement(θ)=π/8 radians=180/8=45/2=22.50Angular…

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A: we have v=ωA2-x2vmax=Aω⇒v=vmaxAA2-x2=vmax1-x2A2

Q: In the experiment, you will study an oscillator called a "torsion pendulum." In this case, the…

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