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**SHORT ANSWER TYPE QUESTIONS**

**Q.1. Find the time period of mass M when displaced from its equilibrium positon and then released for the system shown in Figure.**

**Ans. **For observing oscillation, we have to displace the block slightly beyond equilibrium position and find the acceleration due to the restoring force.

Let in the equilibrium position, the spring has extended by an amount x_{0}.

Tension in the spring = kx_{0}

For equilibrium of the mass M, Mg = 2kx_{0}

Let the mass be pulled through a distance y and then released. But, string is inextensible, hence the spring alone will contribute the total extension y + y = 2y, to lower the mass down by y from initial equilibrium mean position x_{0}. So, net extension in the spring (x_{0} + 2y). From F.B.D of the block,

2k (x_{0} + 2y) - Mg = Ma

2kx_{0} + 4ky - Mg = Ma ⇒ Ma = 4ky

K and M being constant.

∴ a ∝ -x. Hence, motion is SHM

Comparing the above acceleration expression with standard SHM equation a = -ω^{2}x, we get**Q.2. Show that the motion of a particle represented by y = sin ω t - cos ω t is simple harmonic with a period of 2π/ω.Ans.** A function will represents S.H.M. if it can be written uniquely in the form of a or a sin (2π/T (t+ϕ))

Now y = sinωt - cosωt

Comparing with standard SHM y = a s sin (2π/T(t+ϕ))

We get w = 2π/T or T = 2π/ω.

Ans.

The potential energy of the oscillator at this position,

...(i)

Maximum energy of the oscillator = Maximum potential energy = Total energy

TE = 1/2 mω

where, A = amplitude of motion

We are given, PE = 1/2 TE

⇒

⇒

Ans.

= -U

= -U

= -U

We know that F = -kx

So, k = U

Ans.

Ans.

θ

For the first, θ = 2°, ∴ sin (ωt + δ

For the 2nd, θ = -1°, ∴ sin (ωt + δ

∴ ωt + δ

∴ δ

**LONG ANSWER TYPE QUESTIONS**

**Q.1. A person normally weighing 50 kg stands on a massless platform which oscillates up and down harmonically at a frequency of 2.0 s ^{-1} and an amplitude 5.0 cm. A weighing machine on the platform gives the persons weight against time.(a) Will there be any change in weight of the body, during the oscillation?(b) If answer to part (a) is yes, what will be the maximum and minimum reading in the machine and at which position?Ans.** (a) Weight in weight machine will be due to the normal reaction (N) by platform. Consider the top position of platform, two forces due to weight of person and oscillator acts both downward.

So motion is downward. Let with acceleration a then

ma = mg - N ...(i)

When platform lifts form its lowest position to upward

ma = N - mg ...(ii)

a = ω^{2}A acceleration of oscillator

(i) ∴ Form (i) equation

N = mg - mω^{2}A

Where A is amplitude, ω angular frequency, m mass of oscillator.

ω = 2πv = 2π x 2 = 4π rad/sec.

A = 5 cm = 5 x 10^{-2} m, m = 50 kg

N = 50 x 9.8 - 50 x 4π x 4π x 5 x 10^{-2}

= 50[9.8 - 16π^{2} x 5 x 10^{-2}]

= 50[9.8 - 80 x 3.14 x 3.14 x 10^{-2}]

N = 50[9.8 - 7.89] = 50 x 1.91 = 95.50 N

So minimum weight is 95.50 N.

(ii) form (ii) N - mg = ma

For upward motion form lowest point of oscillator.

N = mg + ma = m(a+g)

= m[9.81 + ω^{2}A] a = ω^{2}A

= 50[9.81 + 16π^{2} x 5 x 10^{-2}]

= 50[9.81 + 7.89] = 50 [ 17.70]

N = 885.00 N

(a) Hence, there is a change in weight of the body during oscillation.

(b) The maximum weight is 885 N, when platform moves from lowest to upward direction.

And the minimum is 95.5 N, when platform moves from highest point to downward direction.**Q.2. A body of mass m is attached to one end of a massless spring which is suspended vertically from a fixed point. The mass is held in hand so that the spring is neither stretched nor compressed. Suddenly the support of the hand is removed. The lowest position attained by the mass during oscillation is 4 cm below the point, where it was held in hand.(a) What is the amplitude of oscillation?(b) Find the frequency of oscillation?Ans.** (a) when mass m is held in support by hand the extension in spring will be zero as no deforming force acts on spring.

Let the mass reaches at its new position unit displacement from previous.

Then P.E. of spring or mass = gravitational P.E. lost by man

P.E = mg x

But P.E. due to spring is 1/2 kx^{2}k = ω^{2}A

∴ 1/2 kx^{2} = mg x

x = 2mg/k

Mean position of spring by block will be when let extension is x_{0} then

F = +kx_{0}

F= mg ∴ mg = +kx_{0} or x_{0} = mg/k ...(ii)

From (i) and (ii)

x = 4 cm ∴ 4 = 2x_{0}

x_{0} = 2 cm.

The amplitude of oscillator is the maximum distance from mean position i.e., x - x_{0} = 4 - 2 = 2 cm.

(b) Time period T = which does not depend on amplitude

2mg/k = x from (i)

Oscillator will not rise above the positive from where it was released because total extension in spring is 4 cm when released and amplitude is 2 cm. So it oscillates below the released position.**Q.3. A cylindrical log of wood of height h and area of cross-section A floats in water. It is pressed and then released. Show that the log would execute S.H.M. with a time period.where m is mass of the body and ρ is density of the liquid.Ans.** When log is pressed downward into the liquid then and upward Buoyant force (B.F.) action it which moves the block upward and due to inertia it moves upward from its mean position due to inertia and then again come down due to gravity. So net restoring force on block = Buoyant force – mg

V = volume of liquid displaced by block

Let when floats then

mg = B.F. or mg = V ρg

mg = Ax_{0}ρg ...(i)

A = area of cross-section

x_{0} = Height of block liquid

Let x height again dip in liquid when pressed into water total height of block in water = (x + x_{0})

So net restoring force = [A(x+x_{0})] ρ.g - mg

F_{restoring} = A x_{0}ρg + Axρg - Ax_{0}ρg

F_{restoring} = - Axpg

(as Buoyant force is upward and x is downward)

F_{restoring }= ∝ - x

So motion is SHM here k Aρg

F_{restoring} = -Apgx

ma = - Apgx

k = Aρg**Q.4. One end of a V-tube containing mercury is connected to a suction pump and the other end to atmosphere. The two arms of the tube are inclined to horizontal at an angle of 45° each. A small pressure difference is created between two columns when the suction pump is removed. Will the column of mercury in V-tube execute simple harmonic motion? Neglect capillary and viscous forces. Find the time period of oscillation.Ans.** Consider the liquid in the length dx . It’s mass is A ρdx at a height x.

PE = A ρdx gx

The PE of the left column

Similarly, P.E. of the right column

h_{1} = h_{2} = l sin 45° where l is the length of the liquid in one arm of the tube.

Total P.E. = Aρgh^{2} = Aρgl^{2} sin^{2} 45°

If the change in liquid level along the tube in left side in y, then length of the liquid in left side is l – y and in the right side is l + y.

Total P.E. = Aρg(l – y)^{2} sin^{2} 45° + Aρg(l + y)^{2} sin^{2} 45°

Change in PE = (PE)_{f} – (PE)_{i}

= Aρg [y^{2} + l^{2}]

Change in K.E. = 1/2 Aρ2ly^{2}

Change in total energy = 0

Differentiating both sides w.r.t. time,

2Aρgy + 2 Aρly = 0

ly + gy = 0

y + g/l (y) = 0

ω^{2} = g/l**Q.5. A tunnel is dug through the centre of the Earth. Show that a body of mass ‘m’ when dropped from rest from one end of the tunnel will execute simple harmonic motion.Ans. **Acceleration due to gravity at P = g.x/R, where g is the acceleration at the surface.

Force = mgx/R = -k.x, k = mg/R

Motion will be SHM with time period

**Ans.** Assume that t = 0 when θ = θ_{0}. Then,

θ = θ_{0} cos ωt

Given a seconds pendulum ω = 2π

At time t_{1}, let θ = θ_{0}/2

∴

θ = -θ_{0}2π sin 2πt

At t_{1} = 1/6

θ

Thus the linear velocity is

perpendicular to the string.

The vertical component is

u_{y} = -√3πθ_{0}lsinθ_{0}

and the horizontal component is

u_{x} = -√3πθ_{0}lcosθ_{0}

At the time it snaps, the vertical height is

H′ = H + l (1 – cos(θ_{0}/2))

Let the time required for fall be t, then

H' = u_{y}t + (1/2)gt^{2} (notice g is also in the negative direction)

Neglecting terms of order θ^{2}_{0} and higher,

Now H' ≃ H + l (1 – 1) = H ∴ t ≈

The distance travelled in the x direction is u_{x}t to the left of where it snapped.

To order of θ_{0},

At the time of snapping, the bob was

l sinθ_{0}≃lθ_{0} distance from A.

Thus, the distance from A is

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