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A harmonic wave function is a periodic function whose functional form is sine or cosine.

Step 1. \(v=\omega\sqrt{a^2\,-\,y^2},\)

\(v_2=\omega\sqrt{a^2\,-\,y_1^2}\)

And \(v_2 =\omega\sqrt{a^2\,-y^2_2}\)

4 = \(\omega\sqrt{a^2\,-\,3^2}\) ...(i)

And  3 = \(\omega\sqrt{a^2\,-\,4^2}\) ...(ii)

Dividing eqn (i) by (ii) :

\(\frac{4}{3}=\frac{\sqrt{a^2-9}}{\sqrt{a^2-16}}\)

i.e., \(\frac{16}{9}=\frac{\sqrt{a^2-9}}{\sqrt{a^2-16}}\)

Or 16a² − 256 = 9a² − 81

Or a²= 25

Or a = 5 m.

Step 2. Putting a = 5 in eqn (i),

4 = 4ω

Or ω = 1 rad s^-1 

Step 3. Time taken to travel half amplitude from positive extreme positive extreme position is given by displacement,

x =acos ωt

i.e.,  \(\frac{5}{2}= 5cos(1\times t) \,or\,cos\, t=\frac{1}{2}\)

Or  t = cos^-1\(\frac{1}{2}\) = 60º = \(\frac{\pi}{3}\)

Or  t = \(\frac{3.142}{3}\) = 1.047 s.

(a) It represents S.H.M. of period T = 2μ/ω.

(b) It represents periodic motion of period, T = 2π/ω, but it is not a S.H.M.

(c) It represents a S.H.M. of period 2π/2ω i.e., π/ω.

(d) It represents a periodic motion of period 2π ω, but it is not a S.H.M.

(e) It represents a non-periodic motion.

(f) It represents a non-periodic motion.

(i) The Fig. (a) does not represent periodic motion, as the motion neither repeats nor comes to mean position.

(ii) The Fig. (b) represents the periodic motion, with period equal to 2s.

(iii) The Fig. (c) does not represent periodic motion, because it is not identically repeated.

(iv) The Fig. (d) represents periodic motion, because it is not identically repeated.

Only the relation (c) represents the simple harmonic motion.

Length of the scale = 20 cm = 0.20 m

Total scale reading = 50 kg

Scale reading per unit length of the scale = 50/0.20 = 250 kg m^-1

Now, the time-period of oscillation, T = 0.6 s

Hence, T = 2π√{l/g} ⇒ l = {T²g}/{4π²}

= {(0.6)² x 9.8}/{4 x 9.87} = 0.089 m

Reading on the scale corresponding to the length l is

= 250 x 0.089 kg

= 22.34 kg

= 22.34 x 9.8 N = 219 N

The time period of oscillations is given by,

T = 2π√{(m)/(k₁ + k₂)}

⇒ m = {T²(k₁ + k₂)}/{4π²}    ...(i)

Here, T = 1.5 s, k₁ = k₂ = K(let) and m = 12 kg

Hence, 12 = {(1.5)² (k + K)}/{4 x 9.87}

⇒ ={2 x 2.25 K}/{4 x 9.87} = 12

or, k = {12 x 4 x 9.87}/{2 x 2.25} = 1.05 x 10² Nm^-1

when additional mass of 'm' kg is placed on the tray of mass 12 kg, then 'm' = m + 12 kg and T = 3 sec

Using equation (i), we get

m + 12 = {T²(k₁ + k₂)}/{4π²}

= {T² x 2k}/{4π²}

= {(3)² x 2 x 1.05 x (10)²}/{4 x 9.87}

= 47.87

or, m = 47.87 - 12

= 35.87 kg

y = 4 cos² (t/2) sin x (1000 t)

= 2(1 + cost) sin (1000t)(2 cos²θ = 1 + cos 2θ)

= 2 sin 1000 t + 2 sin 100 t x cos t

= 2 sin 1000 t + (sin 1000 + 1)t + sin (1000 - 1)t

[2 sin A cos A = sin(A + B) + sin(A - B)]

= 2 sin 1000 t + sin 1001 t + sin 999t

= sin 1000 t + sin 1000 t + sin 1001t + sin 999 t

It shows that the given expression is composed of 4 SHMs out of which two are same. This means that the given expression is the result of three independent harmonics.

The two prongs of tuning fork set each other in resonant vibrations and help to maintain the vibrations for a longer time.

Let frequency of I^st tunning fork = x 

frequency of II^nd tunning fork = x + 4 

frequency of III^rd tunning fork = x + 2 (4) 

frequency of IV^th tunning fork = x + 3 (4) 

∴ Let frequency of 24^th tunning fork = x + 23 (4)

octave means, (twice in freq.)

∴  freq. of 24th = 2 × freq. of I^st = 2x

∴ 2x = x + 23 (4) ⇒ x = 92

freq. of 24^th = 2 × 92 = 184 H₃.

(c) simple harmonic and time period is independent of the density of the liquid.

Given, m = mass of cylinder

h = height of cylinder

h₁ = length of cylinder dipping in liquid in equilibrium position

ρ = density of liquid

A = area of cross section of cylinder

mg = buoyant force

= weight of water displaced by body

= ρ(Ah₁)g …(i)

log is pressed gently through small distance x vertically and released.

F_B = ρA(h₁ + x)g

∴ Net restoring force, F = Buoyant Force – weight

= ρA(h₁ + x)g − mg

= ρA(h₁ + x)g − ρ(Ah₁ )g [from (ii)]

= (Aρg)x 

∴ F and x are in opposite direction.

F = −(Aρg)x

a = \(\frac{-(Aρg)}{m}x\) …(ii)

for standard SHM a = w²x …(iii) 

∴ by (ii) & (iii) w² = \(\frac{Aρg}{m}\) 

or w = \(\sqrt{\frac{Aρg}{m}}\) 

∴ T = \(2\pi\sqrt{\frac{m}{Aρg}}\)

Two times in one vibration, KE and PE become maximum.

Correct answer is (a) – Akx