Physics NCERT Exemplar Solutions Class 11th Chapter Fourteen

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(a, c, d) a) when the particle is 3 cm away from A going towards B, velocity is towards AB.i.e positive. SHM towards mean position so positive.

b) When the particle is at C velocity is towards B hence positive.

c) When the particle is 4 cm away from B is going towards A velocity is negative and acceleration towards mean position. Hence negative.

d) Acceleration is always towards mean position O. when the particle is at B acceleration and force are towards BA that is negative.

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(a, b, d) a) total mechanical energy of the body at any time t is

E= $\frac{1}{2}m$ w²a²

KE=1/2mv²= $\frac{1}{2}m[{\frac{dx}{dt}]}^2$

K_max= ½ mw²a²=E

b) K= ½ mw²a²cos²wt

for a cycle value of coswt is =1/2

 = 1/4 mw²a²= kmax/2

c) v=dx/dt = a coswt

V_maen =V_max+V_min/2

= aw-aw/2=0

d)V_rms= $\sqrt[]{\frac{v_1^2+v_2^2}{2}}=\sqrt[]{\frac{0+a^2{w}^2}{2}}=\frac{aw}{\sqrt[]{2}}$

V_rms=V_max/    $\sqrt[]{2}$

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(a, b, c) At t= 3T/4, the displacement of the particle is zero. Hence particle executing SHM will be at mean position i.e x=0 acceleration is zero and force is also zero.

At t= 4T/3, displacement is maximum i.e extreme position, so acceleration is maximum

At t = T/4 corresponds to mean position, so velocity will be maximum at this position.

At t= T/2 corresponds to extreme position so KE =0 and PE =maximum.

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(a, b, d) x=asinwt

V=dx/dt=awcoswt

A=dv/dt=-aw²sinwt

Force = mass $*acceleration$ = -mw²x

So force is directly proportional to displacement.

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(b), (d) It is clear the curve that points corresponding to t=2s and t=6s are separated by a distance belongs to one time period. hence these points must be in same phase

Similarly points belongs t-1s to t=5sare at separation of one time period. Hence must be in phase.

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(a), (c) the rotation of earth about its axis is periodic because it repeats after a regular interval of time.

The rotation of earth is obviously not a to and fro type of motion about a fixed point. So its motion is SHM.

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(a) V=dy/dt=awcoswt

V_max=aw=30

Acceleration A= $\frac{d{x}^2}{d{t}^2}=-a{w}^{2}sinwt$

A=w²a=60

So after solving w =2rad/s

2 $\frac{\pi }{T}=2rad/s$

T= $\pi sec$

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(c) x=acos (  $\alpha t$ )², so it is SHM motion.

X (t+T)=acos [ $\alpha (t+T)$ ]²

= acos [ $\alpha t^2+\alpha T^2+2\alpha tT$ ]

So it is not periodic.

So it is oscillatory but not periodic