Question 1:
moderateThe equation of a simple harmonic wave is given by $y = 3\sin\frac{\pi}{2}(50t – x)$ where $x$ and $y$ are in metres and $t$ is in seconds. The ratio of maximum particle velocity to the wave velocity is:
(2012 Mains)
Comparing with $y = A\sin(\omega t - kx)$, we get $A = 3$, $\omega = 25\pi$, and $k = \pi/2$. Maximum particle velocity $v_{p} = A\omega$. Wave velocity $v_{w} = \omega/k$. The ratio is $v_{p}/v_{w} = (A\omega) / (\omega/k) = Ak = 3 \times (\pi/2) = 3\pi/2$.
Question 2:
easyTwo waves are represented by the equation $y_1 = a\sin(\omega t + kx + 0.57)text{ m}$ and $y_2 = a\cos(\omega t + kx)text{ m}$, where $x$ is in meter and $t$ in sec. The phase difference between them is
(2011 Pre)
The second wave can be rewritten as $y_2 = a\cos(\omega t + kx) = a\sin(\omega t + kx + \pi/2)$. The phase of $y_2$ is $\phi_2 = \pi/2 \approx 1.57text{ rad}$. The phase of $y_1$ is $\phi_1 = 0.57text{ rad}$. The phase difference is $\Delta\phi = 1.57 - 0.57 = 1.0text{ radian}$.
Question 3:
moderateSound waves travel at $350text{ m/s}$ through a warm air and at $3500text{ m/s}$ through brass. The wavelength of a $700text{ Hz}$ acoustic wave as it enters brass from warm air:
(2011 Pre)
Frequency $f$ remains constant when a wave changes medium. Velocity is given by $v = f\lambda$, which means $\lambda \propto v$. The ratio of velocities is $v_{brass}/v_{air} = 3500/350 = 10$. Thus, the wavelength increases by a factor of 10.
Question 4:
moderateA transverse wave is represented by $y = A\sin(\omega t – kx)$. For what value of the wavelength is the wave velocity equal to the maximum particle velocity?
(2010 Pre)
The maximum particle velocity is $v_{max} = A\omega$. The wave velocity is $v = \omega/k$. Equating the two gives $\omega/k = A\omega$, so $1/k = A$. Since $k = 2\pi/\lambda$, we get $\lambda/(2\pi) = A$, which yields $\lambda = 2\pi A$.
Question 5:
moderateA wave in a string has an amplitude of $2text{ cm}$. The wave travels in the +ve direction of x axis with a speed of $128text{ m/sec}$ and it is noted that 5 complete waves fit in $4text{ m}$ length of the string. The equation describing the wave is
(2009)
Amplitude $A = 2text{ cm} = 0.02text{ m}$. Wavelength $\lambda = 4/5 = 0.8text{ m}$. Wave number $k = 2\pi/0.8 \approx 7.85text{ m}^{-1}$. Angular frequency $\omega = vk = 128 \times 7.85 \approx 1005text{ rad/s}$. For +ve x-direction, $y = A\sin(kx - \omega t) = 0.02\sin(7.85x - 1005t)$.
Question 6:
moderateTwo points are located at a distance of $10text{ m}$ and $15text{ m}$ from the source of oscillation. The period of oscillation is $0.05text{ sec}$ and the velocity of the wave is $300text{ m/sec}$. What is the phase difference between the oscillations of two points?
(2008)
Frequency $f = 1/T = 1/0.05 = 20text{ Hz}$. Wavelength $\lambda = v/f = 300/20 = 15text{ m}$. The path difference is $\Delta x = 15 - 10 = 5text{ m}$. The phase difference is $\Delta\phi = (2\pi/\lambda)\Delta x = (2\pi/15) \times 5 = 2\pi/3$.
Question 7:
easyThe wave described by $y = 0.25\sin(10\pi x – 2\pi t)$ where $x$ and $y$ are in meters and $t$ in seconds, is a wave travelling along the:
(2008)
Comparing with $y = A\sin(kx - \omega t)$, the negative sign indicates it travels in the +ve x direction. Amplitude $A = 0.25text{ m}$. Wave number $k = 10\pi$, so $\lambda = 2\pi/k = 0.2text{ m}$. Angular frequency $\omega = 2\pi$, so $f = \omega/(2\pi) = 1text{ Hz}$.
Question 8:
moderateA transverse wave propagating along x-axis is represented by $y(x,t) = 8.0\sin(0.5\pi x – 4\pi t – \pi/4)$ where $x$ is in metres and $t$ is in seconds. The speed of the wave is:
(2006)
From the given wave equation $y = A\sin(kx - \omega t - \phi)$, we identify $k = 0.5\pi\text{ m}^{-1}$ and $\omega = 4\pi\text{ rad/s}$. The wave speed is given by $v = \omega/k = 4\pi / 0.5\pi = 8\text{ m/s}$.
Question 9:
easyWhich one of the following statement is true?
(2006)
Light waves are electromagnetic in nature and are transverse waves; they do not require a medium to propagate. Sound waves are mechanical waves and are longitudinal in fluids like air.
Question 10:
easyA point source emits sound equally in all directions in a non-absorbing medium. Two points P and Q are at distance of $2text{ m}$ and $3text{ m}$ respectively from the source. The ratio of the intensities of the waves at P and Q is:
(2005)
For a point source in a non-absorbing medium, the intensity $I$ is inversely proportional to the square of the distance $r$ from the source ($I \propto 1/r^2$). Thus, $I_P/I_Q = (r_Q/r_P)^2 = (3/2)^2 = 9/4$.