Search results for "Simple Harmonic Motion"

Physics

Edexcel2024

Gravitational field strength Gm Ideal gas equation g = r 2 pV = NkT Gravitational potential Stefan-Boltzmann law −Gm L = σAT 4 V = grav r L = 4πr2σT 4 Oscillations Wien’s law Simple harmonic motion λ T = 2.898 × 10−3 m K max F = −k x Space a = −ω2x Intensity x = A cos ωt L v = −Aω sin ωt I = 4πd 2 a = ‒Aω2 cos ωt Redshift of electromagnetic radiation

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Physics

CAIE2024

(a) State what is meant by simple harmonic motion. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) A block is suspended from a spring, as shown in Fig. 4.1. spring block h floor Fig. 4.1 The block is pulled down and released at time t = 0. It then oscillates vertically with simple harmonic motion. Fig. 4.2 shows the variation of the velocity v of the block with height h of the base of the block above the floor.

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Physics

CAIE2024

V simple harmonic motion a = – ω 2x velocity of particle in s.h.m. v = v cos ωt

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Physics

CAIE2024

(a) State what is meant by simple harmonic motion. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) A block is suspended from a spring, as shown in Fig. 4.1. spring block h floor Fig. 4.1 The block is pulled down and released at time t = 0. It then oscillates vertically with simple harmonic motion. Fig. 4.2 shows the variation of the velocity v of the block with height h of the base of the block above the floor.

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Physics

CAIE2024

V simple harmonic motion a = – ω 2x velocity of particle in s.h.m. v = v cos ωt

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Physics

CAIE2024

Fig. 5.1 shows a pendulum consisting of a metal sphere suspended by a thin string. thin string metal sphere oscillations Fig. 5.1 (not to scale) The sphere undergoes small oscillations about its equilibrium position. The oscillations may be considered to be simple harmonic. Fig. 5.2 shows the variation with time t of the displacement x of the sphere from its equilibrium position.

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Physics

CAIE2024

V simple harmonic motion a = – ω 2x velocity of particle in s.h.m. v = v cos ωt

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Physics

CAIE2023

V simple harmonic motion a = -ω2x velocity of particle in s.h.m. v = v cosωt o ( ) v = ±ω x2 − x2 o Q electric potential V = 4πε r ο capacitors in series 1/C = 1/C + 1/C + ....

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Physics

Edexcel2023

Gravitational field strength Gm Ideal gas equation g = r 2 pV = NkT Gravitational potential Stefan‑Boltzmann law −Gm L = σAT 4 V = grav r L = 4πr2σT 4 Oscillations Wien’s law Simple harmonic motion λ T = 2.898 × 10−3 m K max F = −k x Space a = −ω2x Intensity x = A cos ωt L v = −Aω sin ωt I = 4πd 2 a = ‒Aω2 cos ωt Redshift of electromagnetic radiation 1 2π T = = Δλ Δf v f ω z = ≈ ≈ λ f c ω = 2π f Cosmological expansion Simple harmonic oscillator v = H d m

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Physics

CAIE2023

(a) The defining equation of simple harmonic motion is a = -ω2x. (i) State the relation between ω and the frequency f. (ii) State the significance of the negative (-) sign in the equation. [2] (b) A frictionless trolley of mass m is held on a horizontal surface by means of two similar springs, each of spring constant k. The springs are attached to fixed points as illustrated in Fig. 2.1. spring trolley fixed point Fig. 2.1 When the trolley is in equilibrium, the extension of each spring is e. The trolley is then displaced a small distance x to the right along the axis of the springs. Both springs remain extended. (i) Show that the magnitude F of the restoring force acting on the trolley is given by F = 2kx. [2] (ii) The trolley is then released. Show that the acceleration a of the trolley is given by −2kx a = m [2] © UCLES 2005 9702/04/SPECIMEN PAPER PPPMMMTTT

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Physics

Edexcel2022

Gravitational field strength Gm Ideal gas equation g = r 2 pV = NkT Gravitational potential Stefan‑Boltzmann law −Gm L = σAT 4 V = grav r L = 4πr2σT 4 Oscillations Wien’s law Simple harmonic motion λ T = 2.898 × 10−3 m K max F = −k x Space a = −ω2x Intensity x = A cos ωt L v = −Aω sin ωt I = 4πd 2 a = ‒Aω2 cos ωt Redshift of electromagnetic radiation

Question

Physics

Edexcel2021

Gravitational field strength Gm Ideal gas equation g = r 2 pV = NkT Gravitational potential Stefan-Boltzmann law −Gm L = σAT 4 V = grav r L = 4πr2σT 4 Oscillations Wien’s law Simple harmonic motion λ T = 2.898 × 10−3 m K max F = −k x Space a = −ω2x Intensity x = A cos ωt L v = −Aω sin ωt I = 4πd 2 a = ‒Aω2 cos ωt Redshift of electromagnetic radiation

Question

Physics

Edexcel2020

Gravitational field strength Gm Ideal gas equation g = r 2 pV = NkT Gravitational potential Stefan-Boltzmann law −Gm L = σAT 4 V = grav r L = 4πr2σT 4 Oscillations Wien’s law Simple harmonic motion λ T = 2.898 × 10−3 m K max F = −k x Space a = −ω2x Intensity x = A cos ωt L v = −Aω sin ωt I = 4πd 2 a = ‒Aω2 cos ωt Redshift of electromagnetic radiation

Question

Physics

CAIE2016

(a) The defi ning equation for simple harmonic motion is a = −ω 2x. (i) State the relation between ω and the frequency f. ...................................................................................................................................... [1] (ii) State the signifi cance of the negative (−) sign in the equation. ...................................................................................................................................... [1] (b) A frictionless trolley of mass m is held on a horizontal surface by means of two similar springs, each of spring constant k. The springs are attached to fi xed points, as illustrated in Fig. 3.1. spring trolley fixed point Fig. 3.1 When the trolley is in equilibrium, the extension of each spring is e. The trolley is then displaced a small distance x to the right along the axis of the springs. Both springs remain extended. (i) Show that the magnitude F of the restoring force acting on the trolley is given by F = 2kx. [2] (ii) The trolley is then released. Show that the acceleration a of the trolley is given by 2kx a = − . m [2] © UCLES 2014 9702/04/SP/16 [Turn over PPMMTT

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Physics

CAIE2016

V simple harmonic motion a = – (cid:90)(cid:3)2x velocity of particle in s.h.m. v = v cos (cid:90)t

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