# KPK 11th Class Physics Chapter 7 Oscillation Short Questions Answers

KPK 11th Class Physics Chapter 7 Oscillation Short Questions with answers are combined for all 11th class(Intermediate/hssc) Level students.Here You can prepare all Physics Chapter 7 Oscillation short question in unique way and also attempt quiz related to this chapter. Just Click on Short Question and below Answer automatically shown. After each question you can give like/dislike to tell other students how its useful for each.

**Class/Subject: 11th Class ****Physics**

**Chapter Name: Oscillation**

**Board: All ****KPK Boards**

- Malakand Board 11th Class Physics Chapter 7 Oscillation short questions Answer
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## KPK 11th Class Physics Chapter 7 Oscillation Short Questions Answers

Micro wave oven:

Oven is an important application of resonance. In micro wave oven, the micro wave with a frequency similar to the natural frequency of vibrations of water molecules or fat molecules of the food are used. When food is placed in the oven, the water molecules and fat molecules of food resonates. As a result, these molecules absorbs energy from micro waves and consequently the food gets heated up.

- Radio Reciever:

Tuning of radio is another important application of resonance. when we turn the knob of a radio to tune a station, we are changing the frequency of the electrical circuit of the receiver, to make it equal to the transmission frequency of radio station. When the two frequency match each other, maximum absorption of energy takes place and thus we hear the concern station very clearly.

We know that the time period of a simple pendulum is given by, T=2π√l/g ………………………………..(1)

Now when the length is doubled, then we have

L’ = 2l ………………………………………(2)

Now for ‘l’ , eq (1) can be written as,

T’ = 2π√ l‘/g ………………………….(3)

Putting eq (2) in eq (3) we get,

T’ = 2π √2l/g

- – = √2 (2π √lg)……………………………(4)

Putting eq (1) in eq (4), we get

T’ = √2 T …………………………(5)

Eq (5) shows that the time of simple pendulum will increases 2 – times if its length is doubled.

We know that the frequency of simple pendulum is given by,

F= 1/π√g/e ……………………..(1)

Now we put, l =1m, g = 9.8 m/sec2 in eq (1), we get,

F = 1/2×3.14√9.8/1

- F = 3.13/6.28 => f= 0.498 Hz
- F = 0.5 Hz

Free oscillation:

When a system oscillate with its own natural frequency and no net force acts on it, then such oscillation of the system is known as free oscillation.

Example:

When a simple pendulum is displaced from its mean position and them left free, it execute oscillations.

Forced oscillation:

When a system oscillate under the influence of a net external force, then such oscillation is known as forced oscillation.

Example:

If the mass of a vibrating pendulum is struck again and again then such oscillations of the pendulum is called forced oscillations.

The vibrations of a factory floor caused by the running of heavy machinery is another example of forced oscillation.

When a simple pendulum is set into vibrations and left untouched, then with the passage of time its amplitude decreases due to air resistance. The energy of oscillating system is utilized against air resistance, due to which damped oscillations are produced. The amplitude of oscillations decreases gradually and finally the pendulum stops at its mean position.

The velocity of a simple harmonic oscillator is given by,

V= W √x^{2}–x^{2} ………………(1)

While the acceleration is given by,

A = –w^{2}x………………….(2)

Now when the body is displaced to extreme position ‘x’ and then position ‘x’. and then released, the body execute S.H.M and its velocity becomes maximum at its mean position. So for maximum velocity we consider ‘x’ and put x=0 so eq (1) becomes,

V_{max}=w √x^{2}–o^{2} = w√x^{2}

- V
_{max}=wx ……………………………..(3)

Now put x = 0 in eq (2), we get,

A = w^{2}×0 => a = 0

Eq (3) and eq (4) shows that in S.H.M, the acceleration is zero when the velocity is maximum.

Yes, there is a deep relation b/e ‘F’ and ‘x’ in mass-spring system. We know that the acceleration of a mass-spring system is given by,

a= -(K/m) x

- ma = -kx => F = -kx …………………..(1)

eq (1) represents the mathematical form of hook’s law for mass-sprin system. In eq (1) ‘k’ is constant and is known as spring constant. Put K=constant in eq (1), we get,

F = constant (-x)

- F ∝ (-x) …………………………(2)

Relation (2) represents the connection b/w ‘F’ and ‘x’ in mass – spring system.

Relation (2) shows that, greater the force applied, greater will be extension produced in the spring and vice versa.

We know that the frequency of a simple pendulum is given by,

F = 1/2π√g/l ………………….(1)

Eq (1) shows that the frequency of simple pendulum depends upon the length of the pendulum. It does not depends on the amplitude of vibrations.

Yes, a singer holding a note of right frequency can shatter a glass due to resonance phenomenon.

Every solid body has a certain natural frequency of vibration. If a singer holds a note of a particular frequency equal to the natural frequency of the glass, then resonance occurs and amplitude of vibration of the glass atoms [molecules] will inverses more and more. As a result, the glass may shatter into pieces.