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The velocity of a particle moving with simple harmonic motion is____ at the mean position.
Equation of SHM particle:
Vmax = aω
So the velocity is maximum at mean position
The periodic time (tp) is given by:
Periodic time is the time taken for one complete revolution of the particle.
∴ Periodic time (tp) = 2 π/ω seconds.
A frequency of 1Hz corresponds to:
Frequency is defined as time taken to perform one oscillation by the object.
Hence, 1Hz corresponds to 1 vibration per sec.
A second pendulum is mounted in a space shuttle. Its period of oscillations will decrease when rocket is:
A particle of mass 10 gm lies in a potential field v = (50x2+100) J/kg. The value of frequency of oscillations in Hz is
If a simple pendulum oscillates with an amplitude 50 mm and time period 2s, then its maximum velocity is:
We know that in a simple harmonic motion the maximum velocity,
Vmax = A⍵
Here A = 50 mm
And ⍵ = 2π / T
= 2π / 2
Hence Vmax = 50 x 10-3.π
= 0.15 m/s
In simple harmonic motion the displacement of a particle from its equilibrium position is given by . Here the phase of motion is
Find the amplitude of the S.H.M whose displacement y in cm is given by equation y= 3 sin157t + 4 cos157t, where t is time in seconds.
When the displacement of a SHM is:
y=a sin wt+ b cos wt
Here, a = 3, b = 4
Amplitude, A= √(32+42) = 5 cm
Hence option B is correct.
What will be the phase difference between bigger pendulum (with time period 5T/4 ) and smaller pendulum (with time period T) after one oscillation of bigger pendulum?
After one oscillation of a bigger pendulum i.e. 5T/4, ¼ of the total phase is travelled by the smaller pendulum while the bigger is still at initial position.
Thus, the phase difference between two is: ¼ (2π) - 0 = π/2
A particle executes linear simple harmonic motion with an amplitude of 2 cm. When the particle is at 1 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is:
∴ We get, ω = √3 s-1
T = 2π / √3