Amplitude of Oscillation!

In the figure mA=mB=1kgm_A=m_B=1kg. Block AA is neutral while qB=1Cq_B=-1C. sizes of AA and BB are negligible. BB is released from rest at a distance 1.8m1.8m from AA. Initially springs is neither compressed nor elongated. The amplitude of oscillation of the combined mass will be?

Note by Advitiya Brijesh
6 years, 3 months ago

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Block BB experiences a force of F=qBE=10 NF = q_B E = -10 \ \text{N} over a distance of Δx=1.8 m\Delta x = -1.8 \ \text{m}.

So, it's velocity just before it hits Block AA is vB=2FΔxmB=6 m/sv_B = -\sqrt{\dfrac{2 F \Delta x}{m_B}} = -6 \ \text{m/s}.

Once Block AA and Block BB collide, their combined mass is mA+mBm_A+m_B.

By conservation of momentum, the velocity of the combined mass just after collision is: v0=mBmA+mBvB=3 m/sv_0 = \dfrac{m_B}{m_A+m_B}v_B = -3 \ \text{m/s}.

The total force acting on Block AA and Block BB is kx+qBE-kx+q_B E (spring force plus electric field force).

Using Newton's 2nd Law, the position of the combined mass as a function of time is given by the differential equation

(mA+mB)x=kx+qBE (m_A+m_B)x'' = -kx + q_B E, i.e. (2 kg)x+(18 N/m)x+10 N=0(2 \ \text{kg})x'' + (18 \ \text{N/m})x + 10 \ \text{N} = 0,

with initial conditions x(0)=0 mx(0) = 0 \ \text{m} and x(0)=3 m/sx'(0) = -3 \ \text{m/s}.

The solution is x(t)=(0.556 m)cos((3 rad/s)t)(1 m)sin((3 rad/s)t)(0.556 m)x(t) = (0.556 \ \text{m})\cos\left((3 \ \text{rad/s})t\right) - \left(1 \ \text{m}\right)\sin\left((3 \ \text{rad/s})t\right) - (0.556 \ \text{m}).

The amplitude of oscillation is simply (0.556 m)2+(1 m)2=1.144 m\sqrt{(0.556 \ \text{m})^2 + (-1 \ \text{m})^2} = 1.144 \ \text{m}.

Let me know if there are any mistakes.

Jimmy Kariznov - 6 years, 3 months ago

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thanks! I did it using W.E. Theorem.

Advitiya Brijesh - 6 years, 3 months ago

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You have probably done no mistake. Here is an easier one: Equillibrium position is x=59,vx=0=3m/sx = \frac{-5}{9} , v_{x = 0} = 3m/s Applying conservation of energy from x=0tox=59 x = 0 to x = \frac{-5}{9} , 12(mA+mB)((vmax)2(vx=0)2)+12kx2=Ex\frac{1}{2} (m_{A} + m_{B})((v_{max})^{2} - (v_{x=0})^{2}) +\frac{1}{2}kx^{2} = Ex Solving, we get vmax=1063 v_{max} = \frac{\sqrt{106}}{3} , Now, A=vmaxw=10633A = \frac{v_{max}}{w} = \frac{\frac{\sqrt{106}}{3}}{3} [as w = angular frequency =3radsec 3 \frac{rad}{sec} ] = 1.144m1.144 m

jatin yadav - 6 years, 3 months ago

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Can you please tell how to write in different paragraph, even the lines in different paras appear together.

jatin yadav - 6 years, 3 months ago

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You have not mentioned anything about the nature of collision i.e. the value of coefficient of restitution for the collision to take place. Anyway, assuming it is a perfectly elastic collision, find the energy of the system just before collision and at either of compressed or elongated states. From there get the compression or elongation and your amplitude is double that value. ( The value of k is not visible enough, looks to be 16N/m , amplitude comes out to be 3m)

Nishant Sharma - 6 years, 3 months ago

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The collision is perfectly inelastic, and K=18N/mK=18 N/m

Advitiya Brijesh - 6 years, 3 months ago

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What is projectiles? A projectile is fired at an angle of 60 degre,to the horizontal and with the initial velocity of 80m/s.calculate A) time of flight B) the time taken to get to the maximum hight C) the rane D) the maximum range E) the velocity projection 2seconds after being fired(g=10m/s

Efomo Odia - 6 years, 3 months ago

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http://en.wikipedia.org/wiki/Projectile_motion

Advitiya Brijesh - 6 years, 3 months ago

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