# Phasors

## Trending Questions

**Q.**A particle starts from point x=−√32A and move towards negative extreme as shown in the figure below. If the time period of oscillation is T, then

- The equation of the SHM is x=Asin(2πTt+4π3).
- The time taken by the particle to go directly from its initial position to negative extreme is T12.
- The time taken by the particle to reach mean position is T3.
- The equation of the SHM is x=Asin(2πTt+π3).

**Q.**Figure shows the circular motion of a particle. The radius of the circle, the time period T, sense of revolution and the initial position are indicated on the figure. The simple harmonic motion of the x-projection of the radius vector of the rotating particle P is

- x(t)=B sin(2πt30)
- x(t)=B cos(πt15)
- x(t)=B sin(πt15+π2)
- x(t)=B cos(πt15+π2)

**Q.**A SHM is represented by the equation x=10 sin(πt+π6) in SI units. Find the maximum velocity of particle executing SHM.

- 10π m/s
- 10πm/s
- −10 π m/s
- 2π3m/s

**Q.**A particle executes linear SHM of amplitude A along x− axis. At t=0 the position of the particle is x=A2 and it moves along +x direction. Find the phase constant (δ) if the equation is written as x=A sin(ωt+δ)

- 60∘
- 30∘
- 120∘
- 150∘

**Q.**

Figure (15-E11) shows a string stretched by a block going over a pulleys. The string vibrates in its tenth harmonic in unison with a particular tuning fork. When a beaker containing water is brought under the block so that the block is completely dipped into the beaker, the string vibrates in its eleventh harmonic. Find the desnity of the material of the block.

**Q.**In SHM, the distance of particle from middle point of its path at three consecutive seconds are found to be x, y and z. Then time period of SHM is:

- 2πcos−1(x+z2y)
- 2πtan−1(x+yz)
- 2πsin−1(x+z2y)
- 2πcos−1(x+z2y)

**Q.**A particle performs SHM given by the equation x=3cosωt+4sinωt. Find the phase constant and amplitude.

- 50∘, 5 units
- 37∘, 3.5 units
- 53∘, 3.5 units
- 37∘, 5 units

**Q.**

Three particles starts from origin at tha same time . One along positive x-axis with velocity v_{1} second along positive y axis with velocity v_{2} and third along the line y=x in such a speed that all three particles will always remain in a straight line , what is the velocity of third particle.

**Q.**

The equation of a simple harmonic motion is X = 0.34 cos ( 3000 t + 0.74 ) where *X* and *t* are in *mm* and *sec*. The frequency of motion is

3000/2π

0.74/2π

3000/π

3000

**Q.**The displacement-time(x−t) graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at t=43s is

- √332π2 cm/s2
- −√332π2 cm/s2
- −π232 cm/s2
- π232 cm/s2

**Q.**Which of the following equations represents a simple harmonic motion? The displacement of the system from the equilibrium position is x at time t and α is a positive constant.

- = 0

**Q.**

The equation y=4sin(pi*t/2 + pi/3) represents the displacement of a particle executing SHM, the time period of the SHM is?

**Q.**A particle is subjected to two SHMs of displacement x1=5sin(ωt) and x2=3sin(ωt+60∘) respectively. Find out the equation of displacement of the resulting SHM.

- xnet=7sin(ωt+tan−(5√311))
- xnet=7sin(ωt+60∘+tan−(5√311))
- xnet=7sin(ωt+tan−(3√313))
- xnet=7sin(ωt+60∘+tan−(3√313))

**Q.**Two particles execute SHM of same amplitude of 20 cm with same period along the same line about the same equilibrium position .The maximum distance between two is 20 cm .then their phase difference in radians is ?

**Q.**Assertion: Time taken by a particle executing SHM of amplitude A to move from x=A to x=√3A2 is same as the time taken by the particle to move from x=√3A2 to x=A2.

Reason: Corresponding angles rotated in the reference circle are same in the given time intervals.

- If both assertion and reason are true and the reason is correct explanation of the assertion.
- If both assertion and reason are true but reason is not the correct explantion of assertion.
- If assertion is false but the reason is true.
- If assertion is true but the reason is false.

**Q.**Assertion :A particle, simultaneously subjected to two simple harmonic motions of same frequency and same amplitude, will perform SHM only if the two SHM's are in the same direction. Reason: A particle, simultaneously subjected to two simple harmonic motions of same frequency and same amplitude, perpendicular to each other the particle will be in uniform circular motion.

- Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
- Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
- Assertion is correct but Reason is incorrect
- Assertion is incorrect and Reason is incorrect

**Q.**The potential energy of a particle of mass 'm' situated in a uni-dimensional potential field varies as U(x)=U0[1−cosax] where U0 and a are constants. The time period of small oscillations of the particle about the mean position:

- 2π√maU0
- 2π√amU0
- 2π√ma2U0
- 2π√a2mU0

**Q.**A straight conducting rod PQ is executing SHM in xy−plane from x=−d to x=+d. Its mean position is x=0 and its length is along y− axis. There exists a uniform magnetic field B from x=−d to x=0 pointing inward normal to the paper and from x=0 to =+d there exists another uniform magnetic field of same magnitude B but pointing outward normal to the plane of the paper. At the instant t=0, the rod is at x=0 and moving to the right. The induced emf (ε) across the rod PQ vs time (t) graph will be

**Q.**A particle executing simple harmonic motion along y – axis has its motion describe dy the equation y=Asin(ωt)+B. The amplitude of the simple harmonic motion is

- A
- B
- A + B

**Q.**

What is the conclusion for an object moving with a uniform velocity?

**Q.**The displacement-time(x−t) graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at t=43s is

- √332π2 cm/s2
- −π232 cm/s2
- π232 cm/s2
- −√332π2 cm/s2

**Q.**Two equation of two simple harmonic motion are x=asin(ωt−α) and y=bcos(ωt−α). The phase difference between the two is:-

- 0∘
- α∘
- 90∘
- 180∘

**Q.**The wire of a sonometer has a length of 1m and mass 5×10−4kg. It has a tension of 20N. If the wire is pulled at a point 25cm away from one end and released, the frequency of its vibrations will be

- 200Hz
- 150Hz
- 250Hz
- 100Hz

**Q.**A ring is suspended at a point on its rim and it behaves as a second's pendulum when it oscillates such that its center move in its own plane.The radius of the ring would be (g=π2)

- 0.5m
- 1.0m
- 0.67m
- 1.5m

**Q.**Two blocks are connected by a cord passing over a small frictionless pulley and resting on frictionless planes as shown in the figure. The acceleration of the blocks is:

- 0.33m/s2
- 1m/s2
- 0.66m/s2
- 1.33m/s2

**Q.**A string 1m long is drawn by a 300Hz vibrator attached to its end. The string vibrates in 3 segments. The speed of transverse waves in the string is equal to

- 100 m/s
- 300 m/s
- 200 m/s
- 400 m/s

**Q.**Two points are located at a distance of 10 m and15 m from the source of oscillation. The period of oscillation is 0.05 sec and the velocity of the

wave is 300 m/s. What is the phase difference between the oscillations of two points?

- π
- π/6
- π/3
- 2π/3

**Q.**If the displacement equation of a particle from its mean position is given as y=0.2sin(10πt+1.5π)cos(10πt+1.5π) then, the motion of particle is

- SHM with period 0.2 s
- Periodic but not SHM
- Non-periodic
- SHM with period 0.1 s

**Q.**A uniform chain of length l and mass m lies on a smooth horizontal table with its length perpendicular to the edge of the table and small part overhanging. The chain starts sliding down from rest due to the weight of hanging part. The acceleration and velocity of the chain when length of the hanging portion is x

- gxl, √gx2l
- gxl, √g(l−x)
- gxl, √gx
- gxl, √gl

**Q.**The displacement of a particle varies according to the relation x=4(cosπt+sinπt). The amplitude of the particle is:

- −4
- 4
- 8
- 4√2