The amplitude of a damped oscillator decreases to 0.9 times its original magnitude is 5s. In another 10s it will decrease to α times its original magnitude, where α equals
0.7
0.81
0.729
0.6
C.
0.729
Amplitude of damped oscillator
A uniform cylinder of length L and mass M having cross-sectional area A is suspended, with its length vertical, from a fixed point by a massless spring, such that it is half submerged in a liquid of density σ at the equilibrium position. The extension x_{0} of the spring when it is in equilibrium is
C.
At equilibrium ΣF = 0
For a simple pendulum, a graph is plotted between its kinetic energy (KE) and potential energy (PE) against its displacement d. Which one of the following represents these correctly?
A.
During oscillation, the motion of a simple pendulum KE is maximum of mean position where PE is minimum.
A particle performs simple harmonic motion with amplitude A. It's speed is trebled at the instant that it is at a distance 2A/3 from the equilibrium position. The new amplitude of the motion is:
3A
D.
The velocity of a particle executing SHM at any instant is defined as the time rate of change of its displacement at that instant.
Where ω is angular frequency, A is amplitude and x are displacements of a particle.
Suppose that the new amplitude of the modules be A'.
The initial velocity of a particle performs SHM.
Where A, is initial amplitude and ω is angular frequency.
Final velocity,
A pendulum clock loses 12 s a day if the temperature is 408C and gains 4 s a day if the temperature is 208C. The temperature at which the clock will show correct time, and the co-efficient of linear expansion (α) of the metal of the pendulum shaft are respectively:
25 C; α=1.85×10^{−5}/ °C
60 °C; α=1.85×10−4/ °C
30°C; α=1.85×10−3/°C
55°C; α=1.85×10−2/°8C
A.
25 C; α=1.85×10^{−5}/ °C
Time period of pendulum,
Thus, the coefficient of linear expansion in pendulum clock = 1.85 x 10^{-5}/C
The period of oscillation of a simple pendulum is . The measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using wrist watch of 1 s resolution. The accuracy in the determination of g is:
2%
3%
1%
5%
B.
3%
Time period,
thus, changes can be expressed as
According to the question, we can write
Again time period
A particle moves with simple harmonic motion in a straight line. In first τ s, after starting from rest it travels a distance a, and in next τ s it travels 2a, in the same direction, then
amplitude of motion is 3a
time period of oscillations is 8τ
amplitude of motion is 4a
time period of oscillations is 6τ
D.
time period of oscillations is 6τ
A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90 s, 91 s, 95 s and 92 s. If the minimum division in the measuring clock is 1 s, then the reported mean time should be:
A.
Arithmetic mean time of an oscillating simple pendulum
Mean deviation of a simple pendulum
Given, minimum division in the measuring clock, i.e. simple pendulum = 1s. Thus, the reported mean time of an oscillating simple pendulum =
A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of the air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s.
12
8
6
4
C.
6
If a simple pendulum has the significant amplitude (up to a factor of 1/e of original) only in the period between t = Os to t = τs, then τ may be called the average life of the pendulum. When the spherical bob of the pendulum suffers a retardation (due to viscous drag) proportional to its velocity, with 'b' as the constant of proportionality, the average life time of the pendulum is (assuming damping is small) in seconds:
0.693/b
b
1/b
2/b
D.
2/b
For damped harmonic motion,
ma = - kx - mbv
or
ma + mbv +kx = 0
Solution to above equation is
Where amplitude drops exponentially with time
Average time τ is that duration when amplitude drops by 63%. i.e., becomes A_{o}/e
Thus,