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QUESTION: 1

The velocity of a particle moving with simple harmonic motion is____ at the mean position.

Solution:

__Equation of SHM particle:__

Y=a sinωt

V=aω sinωt

V_{max }= aω

So the velocity is maximum at mean position

QUESTION: 2

The periodic time (t_{p}) is given by:

Solution:

Periodic time is the **time taken for one complete revolution of the particle**.

∴ Periodic time (t_{p}) = 2 π/ω seconds.

QUESTION: 3

A frequency of 1Hz corresponds to:

Solution:

Frequency is defined as **time taken to perform one oscillation by the object**.

Hence, 1Hz corresponds to 1 vibration per sec.

QUESTION: 4

A second pendulum is mounted in a space shuttle. Its period of oscillations will decrease when rocket is:

Solution:

**Time Period, T = 2π √(l/g')**where,

l = Length of seconds pendulum

g’ = Apparent Gravity- For the period of oscillations of Seconds Pendulum to decrease, the Apparent gravity (g’) has to increase because:
- Hence, Time Period of oscillations of Seconds Pendulum will decrease when the rocket is
**ascending up with uniform acceleration**.

QUESTION: 5

A particle of mass 10 gm lies in a potential field v = (50x^{2}+100) J/kg. The value of frequency of oscillations in Hz is

Solution:

QUESTION: 6

If a simple pendulum oscillates with an amplitude 50 mm and time period 2s, then its maximum velocity is:

Solution:

We know that in a simple harmonic motion the maximum velocity,

**V _{max} = A⍵**

Here A = 50 mm

And ⍵ = 2π / T

= 2π / 2

= π

Hence V_{max} = 50 x 10^{-3}.π

= 0.15 m/s

QUESTION: 7

In simple harmonic motion the displacement of a particle from its equilibrium position is given by . Here the phase of motion is

Solution:

QUESTION: 8

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.

Solution:

When the displacement of a SHM is:

y=a sin wt+ b cos wt

__Amplitude of the SHM will be:__

**A=√a**^{2}+b^{2}

Here, a = 3, b = 4

Amplitude, A= √(3^{2}+4^{2}) = 5 cm

Hence option B is correct.

QUESTION: 9

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?

Solution:

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

QUESTION: 10

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:

Solution:

∴ **We get, ω = √3 s ^{-1}
T = 2π / √3**

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