A tunnel has been dug through the centre of the earth and a ball is released in it. It executes S.H.M. with time period

42 minutes
1 day
1 hour
84.6 minutes

Solution

\(T=2\pi \sqrt{\frac{R}{g}}=2\pi \sqrt{\frac{64\times 10^{2}}{9.8}}=2\times \frac{22}{7}\times \frac{8\times 10^{3}}{7\times \sqrt{2}}\)

\(=\frac{\sqrt{2}\times 22\times 8\times 1000}{49\times 60}min =84.6\: min\)

A particle moves such that its acceleration ‘a’ is given by a= – bx where x is the displacement from equilibrium position and b is constant. The period of oscillation is

2 π/b
2 π/\(\sqrt{b}\)
\(\sqrt{2\pi b}\)
\(2\sqrt{\pi/b}\)

Solution

\(T=2\pi \sqrt{\frac{displacement}{acceleration}}=2\pi \sqrt{\frac{x}{bx}}=2\pi /\sqrt{b}\)

A particle of mass 1 kg is moving in S.H.M. with an amplitude 0.02 and a frequency of 60 Hz. The maximum force acting on the particle is

144 π²
188 π²
288 π²
None of these

Solution

Max. force = mass × max. acceleration= m4 π² V²a =1 × 4 ×π²× (60)²× 0.02 = 288 π²

An instantaneous displacement of a simple harmonic oscillator is x = A cos (ωt + π/4). Its speed will be maximum at time

π/4 ω
π/2 ω
π/ω
2π/ω

Solution

Velocity,V=\(\frac{dx}{dt}=-A\: \omega \: sin(\omega +\pi /4)\)

Velocity will be maximum, when

t=π/4 ω

The tension in the string of a simple pendulum is

constant
maximum in the extreme position
zero in the mean position
none of these

Solution

Tension is maximum at the mean position.

The equation of SHM of a particle is A + 4π²x = 0 where a is instantaneous linear acceleration at displacement x. The frequency of motion is

1 Hz
4πHz
¹⁄₄Hz
4 Hz

Solution

Comparing A+ω²X=0

ω²=4π²⇒ω=2π,2πv=2πy=1Hz

If the magnitude of displacement is numerically equal to that of acceleration, then the time period is

1 second
π second
2π second
4π second

Solution

\(y=\omega ^{2}y,\omega =1,\frac{4\pi ^{2}}{T^{2}}=1\: or\: T^{2}=4\pi ^{2}\)

or T = 2π second.

A particle executing simple harmonic motion along y-axis has its motion described by the equations in y=A sin(ωt)+B.The amplitude of the simple harmonic motion is

A
B
A + B
\(\sqrt{A+B}\)

Solution

The amplitude is a maximum displacement from the mean position

The graph shown in figure represents

motion of asimple pendulum startingfrom meanposition
motion of a simple pendulum starting from extreme position
simple pendulum describing a horizontal circle
None of these

Solution

At displacement ± a, the velocity is zero. At zero displacement, velocity is maximum

Resonance is an example of

tuning fork
forced vibration
free vibration
damped vibration

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