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Orlov [11]
3 years ago
7

A mass hanging from a spring undergoes vertical simple harmonic motion.1) Where in the motion is the velocity equal to zero?At t

he point where the spring is unstretched.At the highest point in the oscillation.At the lowest point in the oscillation.
Physics
1 answer:
tamaranim1 [39]3 years ago
4 0

Answer:

At the highest point and at the lowest point the velocity of the mass hung on a spring = 0

Explanation:

Simple Harmonic Motion ( S.H.M) : Simple harmonic motion can be defined as a type of motion were a body vibrates or moves to and fro along a straight line under the influence of a force, so that the acceleration of the body towards a fixed point (equilibrium position) is proportional to its distance or displacement from that point. Examples of  bodies undergoing simple harmonic motion are

<em>⇒ The motion of a mass hung on a spring.</em>

<em>⇒ The motion of a simple pendulum</em>

<em>⇒ The motion of a loaded test - tube in a liquid.</em>

Motion of a mass hung on a spring:When a mass is hung to one end spring and other end is firmly clamped to a rigid support.(i)When the mass is in motion, (ii)it pulled down to its lowest point, passes through it equilibrium position (iii) goes to its highest point.

<em>(1) At the lowest and the highest point during the motion of a mass hung a spring, the velocity = 0</em>

<em>(2) At the equilibrium point or unstretched position the velocity is maximum</em>

<em></em>

<em></em>

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\mathbf a(t)=-g\,\mathbf j

\mathbf v(t)=v_{i,x}\,\mathbf i+\left(v_{i,y}-gt\right)\,\mathbf j

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5\,\mathrm m=y_i+\left(19.096\dfrac{\rm m}{\rm s}\right)(6.1\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.1\,\mathrm s)^2

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{v_{f,y}}^2-{v_{i,y}}^2=2a\Delta y

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\mathbf a(t)=\boxed{-g\,\mathbf j}

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