Hint: The diode is forward biased only for the half cycle.
Step 1: Find the waveform for the first half cycle.
When the input voltage is equal to or less than 5 V, the diode will be reverse biased. It will offer high resistance in comparison to resistance (R) in series. Now, the diode appears in an open circuit. The input waveform is then passed to the output terminals. The result with sin wave input is to dip off all positive going portion above 5 V.
Step 2: Find the waveform for the second half cycle.
If the input voltage is more than +5 V, the diode will be conducting as if forward biased offering low resistance in comparison to R. But there will be no voltage in output beyond 5 V as the voltage beyond +5 V will appear across R.
When input voltage is negative, there will be opposition to 5 V battery in p-n junction circuit. Due to it, the reverse bias voltage of the p-n junction decreases and a voltage appears across the output. When input voltage becomes more than -5 V, the diode will be reverse biased. It will offer high resistance in comparison to resistance R in series. Now, the junction diode appears in the open circuit. The input wave form is then passed on to the output terminals.
The output waveform is shown here in fig. (b)
