A body is executing S.H.M. When its displacement from the mean position are 4 cm and 5 cm, it has velocities 10 cm s^–1 and 8 cm s^–1 respectively. Its periodic time is

π/2 s
π s
3 π/2 s
2 π s

Solution

A particle, with restoring force proportional to displacement and resisting force proportional to velocity is subjected to a force F sin ω₀t. If the amplitude of the particle is maximum for ω = ω₁ and the energy of the particle is maximum for ω = ω₂ then

ω₁ = ω₀ and ω₂ ≠ ω₀
ω₁ = ω₀ and ω₂ = ω₀
ω₁ ≠ ω₀ and ω₂ = ω₀
ω₁ ≠ ω₀ and ω₂ ≠ ω₀

Solution

Amplitude resonance takes place at a frequency of external force which is less than the frequency of undamped natural vibrations. Velocity resonance take splace (i.e., maximum energy) when frequency of external periodic force is equal to natural frequency of undamped vibrations.

A straight rod of negligible mass is mounted on a frictionless pivot and masses 2.5 kg and 1 kg are suspended at distances 40 cm and 100 cm respectively from the pivot as shown. The rod is held at an angle θ with the horizontal and released. Then

the rod executes periodic motion about horizontal position after the release
the rod remains stationary after the release.
the rod comes to rest in vertical position with 2.5 kg mass at the lowest point
the rod executes periodic motion about vertical position after the release

Solution

Torque about hinge

2.5 g × 0.40 cos θ– 1g × 1 cos θ = 0

A forced oscillator is acted upon by a force F = F₀ sin ωt.The amplitude of oscillation is given by \(\frac{55}{\sqrt{2\omega ^{2}-36\omega +9}}\).The resonant angular frequency is

2 unit
9 unit
18 unit
36 unit

Solution

At resonance, amplitude of oscillation is maximum

⇒ 2ω²– 36ω + 9 is minimum

⇒ 4ω – 36 = 0 (derivative is zero)

⇒ ω = 9

A body of mass 0.01 kg executes simple harmonic motion about x = 0 under the influence of a force as shown in figure. The time period of S.H.M. is

1.05 s
0.52 s
0.25 s
0.03 s

Solution

Slope of F - x curve = –k = -⁸⁰⁄_0.2 ⇒ k = 400 N/m,

Time period, T = 2π\(\sqrt{\frac{m}{k}}\)= 0.0314 sec.

A point particle of mass 0.1 kg is executing S.H.M. of amplitude of 0.1 m. When the particle passes through the mean position, its kinetic energy is 8 × 10^–3 Joule.Obtain the equation of motion of this particle if this initial phase of oscillation is 45º.

\(y=0.1sin\left ( \pm 4t+\frac{\pi }{4} \right )\)
\(y=0.2sin\left ( \pm 4t+\frac{\pi }{4} \right )\)
\(y=0.2sin\left ( \pm 4t+\frac{\pi }{4} \right )\)
\(y=0.2sin\left ( \pm 2t+\frac{\pi }{4} \right )\)

Solution

A coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency ω.The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time

at the mean position of the platform
for an amplitude of ^g⁄_ω²
for an amplitude of ^g²⁄_ω²
at the highest position of the platform

Solution

For block A to move in S.H.M.

mg – N= mω²x

where xis the distance from mean position For block to leave contact N = 0

⇒ mg = mω²x ⇒ x = ^g⁄_ω²

A body of mass m falls from a height h onto a pan (of negligible mass) of a spring balance as shown. The spring also possesses negligible mass and has spring constant k.Just after striking the pan, the body starts socillatory motion in vertical directioin of amplitude A and energy E. Then

A=^mg⁄_k
\(A=\frac{mg}{k\sqrt{1+\frac{2kh}{mg}}}\)
\(E=mgh+\frac{1}{2}kA^{2}\)
\(E=mgh+\frac{1}{2}kA^{2}\)

Solution

In the figure shown,the spring is light and has a force constant k. The pulley is light and smooth and the strring is light . The suspended block has a mass m. On giving a slight displacement vartically to the block in the downward direction from its equilibrium position the block executes S.H.M. on being released with time period T. Then

\(T=2\pi \sqrt{\frac{m}{k}}\)
\(T=2\pi \sqrt{\frac{m}{2k}}\)
\(T=2\pi \sqrt{\frac{2m}{k}}\)
\(T=4\pi \sqrt{\frac{m}{k}}\)

Solution

Let the extension in the spring be x₀ at equilibrium. If F₀ be the tension in the string then F₀ = kx₀. Further if T₀ is the tension in the thread then T₀= mg and 2T₀ = kx₀.

Let the mass m bed is placed through a slight displacement x downwards. Let the the new tension in the string and spring be T and F respectively.

A circular hoop of radius R is hung over a knife edge. The period of oscillation is equal to that of a simple pendulum of length

R
2R
3R
^3R⁄₂

Solution

\(T=2\pi \sqrt{\frac{2R}{g}}=2\pi \sqrt{\frac{\iota }{g}}\)

UnAttempted

0

Attempted

0

Correct

0

Wrong

0

Complete