# 11th Class Physics Chapter 7 OSCILLATIONS Short Question Answers

## 11th Class Physics Chapter 7 OSCILLATIONS Short Question Answers Below

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Two characteristics of simple harmonic motion are given as below:-

1.acceleration of a vibrating body is directly proportional to the displacement and is always directed towards the mean (or equilibrium) position.

i.e. a∞-x_{
}2.Total energy of particle executing SHM remain conserved

3.SHM can be represented by a single harmonic function of sine or cosine in thr from of equation.

x=xo sin(ῳ+ᶲ)

x=x0cos(ῳt+ᶲ)

or

phase (ᶲ) is a measure of how for the oscillator is away from its mean

position at time t=0

The knowledge of phase enable us to find how for from its mean position the oscillation was at t=0

No, the frequency oscillator is independent of the amplitude of oscillation provided it is small.

As we know that frequency of oscillation of simple pendulum is

F=1/2π∫g/l

This relation shows that frequency does not depend upon the amplitude but it depends upon the length of pendulum and acceleration due to gravity.

Similarly, in the case of mass-spring system , frequency of oscillation of mass is given by

f=1/2π∫k/m

This relation shows that frequency depends upon mass of the body and spring constant ‘k’ but it is independent of the amplitude.

No, the acceleration of harmonic oscillator does not remain constant during its motion.

Given by

aα-x

or a=-(constant)*x

where x is the displacement from the mean position.

Since the displacement changes continuously during SHM, so its acceleration does not remain constant. The value of acceleration at the mean position will be zero because at this position x=0 and its maximum value will be at the extreme position.

The angle which specifies the displacement as well as the direction of motion of the point executing SHM is called phase angle. Thus, it indicates the state of motion of the vibrating the simple harmonic Oscillor is

Mathematically, the phase angle is Expressed as

θ=ῳt_{
}Where ῳ=angular frequency

t=any instant of time.

(b) It is the angle ‘θ’ which the rotating radius OP makes with reference direction OO1 at any instant ‘t’ as shown in the fag.

It does not define angle between maximum displacement and the driving force.

In order to produce resultant SHM by the addition of two simple harmonic motion, following conditions must be fulfilled

(1)Two SHMs must be parallel (i.e. their phase phases must be in the same direction.

(2)Two SHMs must have the same frequency (i.e. period) but different amplitudes.

(3) These two harmonic motion must have constant phase difference.

If two SHMs are given as

x1=A1 sin ῳt+ and x2=A2 sin(ῳt+ᶲ)

Resultant SHMs will be written as

X=X1+X2=A1 sin ῳt+A2 sin (ῳt+ᶲ)

(i)y=A sim (ῳt+ᶲ)

→ →

(II) a -ῳ2x

.(i) y=A sin(ῳt+ᶲ)

This equation represent the displacement of simple harmonic oscillator as a function of time.

Thus, this equation tells that displacement follows a sine curve i.e. varies harmonically.

‘ᶲ’ is initial phase angle which tells us the start of motion. ῳt is the angle subtended in time t with angular frequency ‘ῳ’ starting from initial phase ‘ᶲ’ (ῳt+ᶲ) is the phase angle made with reference direction. ‘y’ is the instantaneous displacement of a particle performing SHM.

‘A’ is the amplitude of the oscillating .

→ →

(ii) a= -ῳ^{2} x

This equation represent the variation of acceleration of S.H. oscillator as a function of displacement.

This equation tells that the acceleration of simple harmonic oscillator is directly proportional to its displacement and its directed towards the main position.

In the above equation , ‘a’ is the acceleration of a particle executing SHM.

‘ῳ’ is the angular frequency of the particle.

‘x’ is in instantaneous displacement of an oscillating particle, from the mean position.

E ………….. =P.E.+K.E

since total energy of SHM remains constant, therefore any decrease in K,E. or P.E. result increase in P.E. or K.E. respectively.

During SHM, in the absence of frictional force, the K.E. and P.E, are interchange continuously from one from to another but the total energy remain constant. At mean position, the energy is totally kinetic, i.e, K.E. is maximum but P.E. is zero. At the extreme position the K.E. is completely change into P.E. , i.e. P.E. becomes maximum but K.E. is zero.

**Frequency of a second, s pendulum:-**

for a second pendulum

Time period =T= 2sec

But the relation between the frequency and the time period is given by

f= 1/T

Therefore, the frequency of a second’s pendulum is given by

F= ½=0.5 vibrations/sec

Or F=05vib-s

^{-1}

[

**Uses of simple pendulum**

(1) The value of ‘g’ can be found by simple pendulum, because both T and l can be directly measured.

(2) The height of a tower can be measured by determining the time period of a pendulum suspend to the top of that tower up to the ground. i.e. l=gT

^{2}/4π

^{2}

(3) We can find the frequency ‘f’ of a vibrating body by simple pendulum.

^{2})?

Mathematically, is can be written as

F

_{r}= -kx

Negative sign shows that the force ‘f’ is negative i.e directed opposite to displacement ‘x’ (towards mean position).

Displacement =X=0.2m

Spring constant = k

Using the formula, F=kx

Or k= f/x=10/0.2

=100/2=50 Nm

^{-2}

Hence, K=50 Nm

^{-1}Ans

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