When an oscillator completes 100 oscillations its amplitude reduces to ¹⁄₃ of its initial value. What will be its amplitude,when it completes 200 oscillations ?

¹⁄₈
²⁄₃
¹⁄₆
¹⁄₉

Solution

A mass is suspended separately by two different springs in successive order then time periods is t₁ and t₂ respectively. It is connected by both springs as shown in fig. then time period is t₀, the correct relation is

\(t_{0}^{2}=t_{1}^{2}+t_{2}^{2}\)
\(t_{0}^{-2}=t_{1}^{-2}+t_{2}^{-2}\)
\(t_{0}^{-1}=t_{1}^{-1}+t_{2}^{-1}\)
t₀=t₁+t₂

Solution

A clock which keeps correct time at 20ºC, is subjected to 40ºC. If coefficient of linear expansion of the pendulum is 12 × 10^–6 per ºC. How much will it gain or loose in time ?

10.3 seconds/day
20.6 seconds/day
5 seconds/day
20 minutes/day

Solution

Two simple pendulums of length 0.5 m and 20 m respectively are given small linear displacement in one direction at the same time.They will again be in the phase when the pendulum of shorter length has completed oscillations[nT₁=(n–1)T₂, where T₁ is time period of shorter length & T₂ be time period of longer wavelength and n are no. of oscillations completed]

Solution

A simple pendulum attached to the roof of a lift has a time period of 2s in a stationary lift. If the lift is allowed to fall freely the frequency of oscillations of pendulum will be

zero
2 Hz
0.5 Hz
infinity

Solution

When lift is falling freely, the effective acceleration due to gravity inside the lift is zero i.e.g' = g – g = 0.Therefore time period will be infinity and frequency is zero

A Second’s pendulum is placed in a space laboratory orbiting around the earth at a height 3 R from the earth’s surface where R is earth’s radius. The time period of the pendulum will be

zero
23
4 sec
infinite

Solution

The second pendulum placed in a space laboratory orbiting around the earth is in a weightlessness state.Hence g = 0 so T = ∞

The spring constant from the adjoining combination of springs is

K
2 K
4 K
5 K/2

Solution

Here all the three springs are connected in parallel to mass m. Hence equivalent spring constant k =K + K + 2 K = 4 K.

A simple spring has length l and force constant K. It is cut into two springs of lengths l₁ and l₂ such that l₁= n l₂(n = an integer). The force constant of spring of length l₁ is

K(1 + n)
(K/n) (1 + n)
K
K/(n + 1)

Solution

Lissajous figure obtained by combining x = a sin ωt and y = a sin (ωt + π/4) will be a/an

ellipse
straight line
circle
parabola

Solution

Two particles are executing S.H.M. of same amplitude and frequency along the same straight line path. They pass each other when going in opposite directions, each time their displacement is half of their amplitude. What is the phase difference between them ?

5 π/6
5 π/6
π/3
π/6

Solution

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