A point performs simple harmonic oscillation of period \(\mathrm{T}\) and the equation of motion is given by; \(x=a \sin (\omega t+\pi / 6)\). After the elapse of what fraction of the time period, the velocity of the point will be equal to half of its maximum velocity?
1. \( \frac{T}{8} \)
2. \( \frac{T}{6} \)
3. \(\frac{T}{3} \)
4. \( \frac{T}{12}\)
A particle executing simple harmonic motion has a kinetic energy of \(K_0 cos^2(\omega t)\). The values of the maximum potential energy and the total energy are, respectively:
1. \(0\) and \(2K_0\)
2. \(\frac{K_0}{2}\) and \(K_0\)
3. \(K_0\) and \(2K_0\)
4. \(K_0\) and \(K_0\)
The radius of the circle, the period of revolution, initial position and direction of revolution are indicated in the figure.
The \(y\)-projection of the radius vector of rotating particle \(P\) will be:
1. \(y(t)=3 \cos \left(\frac{\pi \mathrm{t}}{2}\right)\), where \(y\) in m
2. \(y(t)=-3 \cos 2 \pi t\) , where \(y\) in m
3. \(y(t)=4 \sin \left(\frac{\pi t}{2}\right)\), where \(y\) in m
4. \(y(t)=3 \cos \left(\frac{3 \pi \mathrm{t}}{2}\right) \), where \(y\) in m
A block of mass \(4~\text{kg}\) hangs from a spring of spring constant \(k = 400~\text{N/m}\). The block is pulled down through \(15~\text{cm}\) below the equilibrium position and released. What is its kinetic energy when the block is \(10~\text{cm}\) below the equilibrium position? [Ignore gravity]
1. \(5~\text{J}\)
2. \(2.5~\text{J}\)
3. \(1~\text{J}\)
4. \(1.9~\text{J}\)
The amplitude and the time period in an S.H.M. are 0.5 cm and 0.4 sec respectively. If the initial phase is radian, then the equation of S.H.M. will be:
1.
2.
3.
4.
A mass m is suspended from two springs of spring constant as shown in the figure below. The time period of vertical oscillations of the mass will be
1.
2.
3.
4.
A simple pendulum hanging from the ceiling of a stationary lift has a time period T1. When the lift moves downward with constant velocity, then the time period becomes T2. It can be concluded that:
1. | \(T_2 ~\text{is infinity} \) | 2. | \(\mathrm{T}_2>\mathrm{T}_1 \) |
3. | \(\mathrm{T}_2<\mathrm{T}_1 \) | 4. | \(T_2=T_1\) |
An SHM has an amplitude \(a\) and a time period \(T.\) The maximum velocity will be:
1. \({4a \over T}\)
2. \({2a \over T}\)
3. \({2 \pi \over T}\)
4. \({2a \pi \over T}\)
Two simple harmonic motions of angular frequency 100 rad s -1 and 1000 rad have the same displacement amplitude. The ratio of their maximum acceleration will be:
1. 1:10
2. 1:102
3. 1:103
4. 1:104
The angular velocities of three bodies in simple harmonic motion are with their respective amplitudes as . If all the three bodies have the same mass and maximum velocity, then:
1. | \(A_1 \omega_1=A_2 \omega_2=A_3 \omega_3\) |
2. | \(A_1 \omega_1^2=A_2 \omega_2^2=A_3 \omega_3^2\) |
3. | \(A_1^2 \omega_1=A_2^2 \omega_2=A_3^2 \omega_3\) |
4. | \(A_1^2 \omega_1^2=A_2^2 \omega_2^2=A^2\) |